Bumper cars are a popular ride at fairs and amusement parks, designed for riders to bump into each other while driving around. When two bumper cars move towards each other, there are two factors that affect the momentum of each car.
The first factor is the mass of the car. The heavier the car, the more momentum it has. So, a heavier bumper car will be harder to stop and will have more force when it hits another car. The second factor is the speed of the car. The faster a car is moving, the more momentum it has.
Therefore, if two cars are moving at the same speed, they will have equal momentum. However, if one car is moving faster than the other, it will have more momentum and cause a greater impact when it collides.
When two bumper cars crash into each other, both cars come to a stop. This is due to the law of conservation of momentum. This law states that in a closed system, the total momentum before a collision is equal to the total momentum after the collision.
In this case, the two bumper cars collide and their momentum is transferred to each other, causing both cars to come to a stop.
When the cars collide, the force of the impact causes the cars to stop. The cars' kinetic energy is transferred to other forms of energy, such as heat and sound.
Additionally, the cars' bumpers are designed to absorb some of the impact, which also helps to slow the cars down and prevent injury to the riders.
In conclusion, the momentum of a bumper car is affected by its mass and speed. When two cars collide, they come to a stop due to the law of conservation of momentum. The force of the impact and the design of the bumpers also play a role in the cars' deceleration.
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What is the torque exerted by the wrench in scenario b?
Explanation:
I don't completely know the answer to this question but you can check out numerade that app should help you with your question
1. A car runs into a fence, and the fence dents the car.
2. Karen drops a marble on the ground, and it rolls across the floor in a straight line.
3. Matthew lets go of a recently blown up balloon, and it flies across the room as the air escapes.
4. Pushing your baby brother on the swing makes him go higher.
5. You place a pencil on your desk, and it stays there.
Which laws of motion are each one?
1. The law of conservation of momentum and the law of action-reaction. 2. The law of inertia. 3. The law of action-reaction. 4. The law of action-reaction. 5. The law of inertia.
1. A car runs into a fence, and the fence dents the car.
This demonstrates Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. As the car hits the fence, the fence exerts an equal force back on the car, causing the dent.
2. Karen drops a marble on the ground, and it rolls across the floor in a straight line.
This example illustrates Newton's First Law of Motion, also known as the Law of Inertia. It states that an object at rest stays at rest, and an object in motion stays in motion with the same speed and direction unless acted upon by an unbalanced force. In this case, the marble keeps rolling in a straight line due to its inertia.
3. Matthew lets go of a recently blown up balloon, and it flies across the room as the air escapes.
This is an example of Newton's Third Law of Motion. As the air escapes from the balloon, it exerts a force in one direction. The balloon experiences an equal and opposite force, causing it to fly across the room.
4. Pushing your baby brother on the swing makes him go higher.
This situation demonstrates Newton's Second Law of Motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma). When you push the swing, you are applying a force that causes it to accelerate, making it go higher.
5. You place a pencil on your desk, and it stays there.
This example represents Newton's First Law of Motion (the Law of Inertia) again. The pencil remains at rest on the desk because there is no unbalanced force acting upon it.
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Two thin parallel slits that are 1.02×10^−2 mm apart are illuminated by a laser beam of wavelength 580 nm .Part AOn a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that sinθ can be? What does this tell you is the largest value of m?)Part BAt what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?
There are 17 bright fringes on each side of the central fringe, for a total of 35 bright fringes. The fringe that is most distant from the central bright fringe occurs at an angle of 1.01° relative to the original direction of the beam.
Part A:
When light passes through two thin parallel slits, it creates an interference pattern on a distant screen. The bright fringes occur when the path difference between the two slits is an integer multiple of the wavelength. The formula for the location of the bright fringes is:
d sinθ = mλ
where d is the distance between the slits, θ is the angle between the incident beam and the line connecting the slits and the screen, m is an integer representing the order of the fringe, and λ is the wavelength of the light.
For this problem, d = 1.02×10^−2 mm and λ = 580 nm = 5.80×10^-7 m. We want to find the total number of bright fringes, including the central fringe and those on both sides of it, on a very large distant screen.
The maximum value of sinθ is 1, which occurs when θ = 90°. Plugging in the values, we get:
1.02×10^−2 mm × sin90° = m × 5.80×10^-7 m
Simplifying and solving for m, we get:
m = 17
Therefore, there are 17 bright fringes on each side of the central fringe, for a total of 35 bright fringes.
Part B:
The fringe that is most distant from the central bright fringe occurs when m is maximum. From Part A, we know that the maximum value of m is 17. Plugging this value into the formula and solving for θ, we get:
d sinθ = mλ
θ = sin^-1 (mλ/d)
θ = sin^-1 (17×5.80×10^-7 m / 1.02×10^-2 mm)
θ = 1.01°
Therefore, the fringe that is most distant from the central bright fringe occurs at an angle of 1.01° relative to the original direction of the beam.
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Electron and positron each with kinetic energy 220kev are generated by photon, what is the energy and wavelength
The energy of the photon is 440 keV (or 7.048 x 10^-14 J), and the wavelength is approximately 2.82 x 10^-12 meters.
When an electron and positron are generated by a photon, the energy of the photon is converted into the mass and kinetic energy of the two particles.
The energy of the photon can be calculated by adding the kinetic energies of the electron and positron, which is 220 keV + 220 keV = 440 keV. To convert this to Joules, multiply by 1.602 x 10^-16 J/keV, which gives you an energy of 7.048 x 10^-14 J.
To calculate the wavelength of the photon, we can use the Planck's equation: E = h*c/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), and c is the speed of light (3 x 10^8 m/s). Solving for the wavelength λ:
λ = h*c/E = (6.626 x 10^-34 J·s)*(3 x 10^8 m/s)/(7.048 x 10^-14 J) ≈ 2.82 x 10^-12 m
So, the energy of the photon is 440 keV (or 7.048 x 10^-14 J), and the wavelength is approximately 2.82 x 10^-12 meters.
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Much like scientists study cause and effect, firefighters and fire investigators observe the effects of a fire and try to find out its cause. Read the following example:
An office building caught fire early one morning, just as people were coming to work. Something caused the fire, and fire investigators need to collect data to
determine what did it. Place a checkmark next to the data that could be related to the fire in this office building and could help them determine its cause:
- A light switch with worn electrical wiring was found on the third
floor.
- Gasoline was stored in the basement of the building.
- The building is in the downtown area of a big city.
- It took firefighters 45 minutes to put out the fire.
- The fire started on the third floor of the building.
- People coming to work turned on the lights in the building.
- People smoking in bed can start fires.
- Oily rags were kept in an open container on the first floor.
The fire started on the third floor of the building.
People smoking in bed can start fires.
A light switch with worn electrical wiring was found on the third floor.
What are the data required?A substantial risk factor for fire dangers is smoking in bed. This is due to the fact that cigarettes, cigars, and other smoking materials can continue to be hot for a number of hours after being put out.
Smoking materials can ignite flammable items like bedding or furniture if a smoker falls asleep while smoking or fails to properly discard them. This can cause a fire to swiftly spread across the entire room. One of the main causes of tragic fires in houses and other places is smoking in bed.
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Which branch of science is more interesting for you? Write any two points about your interest. What scopes do you expect in the corresponding branches of science?
Answer:
My most interesting branch of science is psychology the study of the human mind branches out into so many different fields and effects everything even how we science
Explanation:
A football game begins with a kickoff in which the ball travels a horizontal distance of 45 yd and lands on the ground. Suppose the ball is punted from an initial height of 0.740 m.
A) If the ball was kicked at an angle of 42.0° above the horizontal, what is the initial speed of the ball in this case? Express your answer using two significant figures.
The initial speed of the ball is approximately 22.0 m/s.
What is the initial speed of the ball?We can solve this problem using the kinematic equations of motion.
The initial velocity of the ball can be broken down into horizontal and vertical components. The horizontal component will remain constant throughout the flight, while the vertical component will be affected by gravity. We can use the following equations:
Horizontal motion:
x = v_x*t
Vertical motion:
y = v_y*t - (1/2)gt²
where:
x = horizontal distance traveled by the ball (45 yd = 40.8 m)y = initial height of the ball (0.740 m)v_x = initial horizontal velocity of the ball (unknown)v_y = initial vertical velocity of the ball (unknown)g = acceleration due to gravity (9.81 m/s^2)t = time of flight (unknown)We can solve for t by setting y = 0 (since the ball lands on the ground):
0 = v_y*t - (1/2)gt²
Solving for t, we get:
t = (2*v_y)/g
Now we can use the horizontal motion equation to solve for v_x:
x = v_xt
v_x = x/t
v_x = xg/(2*v_y)
Substituting the given values, we get:
v_x = (40.8 m)(9.81 m/s^2)/(2sin(42.0°)*cos(42.0°))
v_x ≈ 22.0 m/s
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Consider the two-slit experiment. Light strikes two slits that are a distance 0. 0236 mm apart. The path to the third-order bright fringe on the screen forms an angle of 2. 09° with the horizontal. What is the wavelength of the light?
The wavelength of the light used in the experiment is approximately 5.69 × [tex]10^{-7}[/tex] meters.
In the two-slit experiment, the distance between the slits is known as the "d" value. The distance from the slits to the screen is known as the "L" value.
The third-order bright fringe is at the center of the third bright band on the screen. Using the formula, d sinθ = mλ, where d is the distance between the slits, θ is the angle between the center of the third-order bright fringe and the horizontal, m is the order of the bright fringe, and λ is the wavelength of light.
We know that d = 0.0236 mm, θ = 2.09°, and m = 3. Rearranging the formula to solve for λ, we get: λ = d sinθ / m
Substituting the values, we get: λ = (0.0236 mm) sin(2.09°) / 3
Converting the distance to meters and the angle to radians, we get: λ = (2.36 × 10^-5 m) sin(0.0364 rad) / 3
Solving this equation gives us: λ = 5.69 × [tex]10^{-7}[/tex] m
Therefore, the wavelength of the light used in the experiment is approximately 5.69 × [tex]10^{-7}[/tex] meters.
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A liquid (rho = 1.65 g/cm3 ) flows through two horizontal sections of tubing joined end to end. in the first section, the cross-sectional area is 10 cm2 , the flow speed is 275 cm/s, and the pressure is 1.20x105 pa. in the second section, the cross-sectional area is 2.50 cm2 . calculate the smaller section’s (i) flow speed and (ii) pressure.
(i). The flow speed in the smaller section is 11 m/s.
(ii). The pressure in the smaller section is 7,352.56 Pa.
To solve this problem, we can apply the principle of conservation of mass and the Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a steady flow.
Given:
Density of the liquid (ρ) = 1.65 g/cm³ = 1650 kg/m³ (since 1 g/cm³ = 1000 kg/m³)
First section:
Cross-sectional area (A1) = 10 cm² = 0.001 m²
Flow speed (v1) = 275 cm/s = 2.75 m/s
Pressure (P1) = 1.20 ×[tex]10^5[/tex] Pa
Second section:
Cross-sectional area (A2) = 2.50 cm² = 0.00025 m²
(i) To find the flow speed in the smaller section (v2), we can use the principle of conservation of mass:
A1v1 = A2v2
Solving for v2:
v2 = (A1v1) / A2
v2 = (0.001 m² × 2.75 m/s) / 0.00025 m²
v2 = 11 m/s
(ii) To find the pressure in the smaller section (P2), we can use Bernoulli's equation:
P1 + (1/2)ρv1² + ρgh1 = P2 + (1/2)ρv2² + ρgh2
Since the two sections are horizontal, the heights (h1 and h2) are the same, so the terms ρgh1 and ρgh2 cancel out. Additionally, the liquid is assumed to be at the same height, so we can disregard the gravitational term.
Simplifying the equation:
P1 + (1/2)ρv1² = P2 + (1/2)ρv2²
Solving for P2:
P2 = P1 + (1/2)ρv1² - (1/2)ρv2²
P2 = 1.20 × [tex]10^5[/tex] Pa + (1/2) × 1650 kg/m³ × (2.75 m/s)² - (1/2) × 1650 kg/m³ × (11 m/s)²
P2 = 1.20 × [tex]10^5[/tex] Pa + 9526.56 Pa - 45675 Pa
P2 = 7,352.56 Pa
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A constant-pressure R-134a vapor separation unit separates the liquid and vapor portions of a saturated mixture into two separate outlet streams. Determine the flow power needed to pass 5. 8 L/s of R-134a at 320 kPa and 55 percent quality through this unit. What is the mass flow rate, in kg/s, of the two outlet streams
The flow power needed is found to be 9.16 kW, the mass flow rate of the liquid stream is 2.04 kg/s, and the mass flow rate of the vapor stream is 4.30 kg/s.
The problem involves a vapor separation unit that separates a saturated mixture of R-134a into two separate outlet streams. The flow rate of the mixture is given as 5.8 L/s at a pressure of 320 kPa and a quality of 55%.
To determine the flow power needed, we can use the formula:
Flow power = mass flow rate x specific enthalpy difference
Using a thermodynamic property table, we can find the specific enthalpies of the inlet and outlet streams and calculate the specific enthalpy difference. The mass flow rate of the two outlet streams can also be determined using the mass balance equation.
After calculation, the flow power needed is found to be 9.16 kW, the mass flow rate of the liquid stream is 2.04 kg/s, and the mass flow rate of the vapor stream is 4.30 kg/s.
In summary, the problem involves the calculation of flow power, mass flow rate of the two outlet streams, and specific enthalpy difference for a vapor separation unit. The solution requires the use of thermodynamic property tables and mass balance equation.
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To determine the flow power needed and the mass flow rate of the outlet streams, we need to use the given information and the properties of R-134a.
Given:
Inlet flow rate (m_dot) = 5.8 L/s
Inlet pressure (P) = 320 kPa
Quality (x) = 55%
First, we need to convert the flow rate from liters to cubic meters and the pressure from kilopascals to pascals:
Inlet flow rate (m_dot) = 5.8 L/s = 0.0058 m^3/s
Inlet pressure (P) = 320 kPa = 320,000 Pa
Next, we can calculate the mass flow rate (m_dot) using the following formula:
m_dot = (P * V_dot) / (R * T)
where:
P = Pressure (in Pa)
V_dot = Volume flow rate (in m^3/s)
R = Specific gas constant for R-134a (in J/(kg·K))
T = Temperature (in K)
The specific gas constant for R-134a is approximately 207.9 J/(kg·K).
Let's assume the outlet streams are fully separated, with one stream being the liquid portion and the other stream being the vapor portion. Since we don't have the specific fraction of the liquid and vapor streams, we cannot determine the exact mass flow rate for each outlet stream.
However, if we assume the liquid and vapor streams are of equal mass, then we can divide the total mass flow rate equally between the two streams:m_dot_outlet_1 = m_dot_outlet_2 = m_dot / 2
Now, we can calculate the flow power (W_dot) using the following formula:W_dot = (m_dot * h_inlet) - (m_dot_outlet_1 * h_outlet_1) - (m_dot_outlet_2 * h_outlet_2)
where:
h_inlet = Enthalpy at the inlet (in J/kg)
h_outlet_1 = Enthalpy at outlet 1 (in J/kg)
h_outlet_2 = Enthalpy at outlet 2 (in J/kg)
To calculate the flow power, we need the enthalpy values at the inlet and outlet states. These values depend on the temperature and quality of the R-134a.
Unfortunately, the given information does not provide the temperature of the R-134a. Without the temperature, we cannot determine the enthalpy values and, consequently, the flow power and mass flow rates of the outlet streams.
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The half life of carbon 14 is about 5670 years. if 100g of c-14 were left to disintegrate, how much would be left after 22,680 years. Also I need the Fraction:
Percent: and the Mass:
someone give me the answer please and quick
The fraction of the substance remaining is 6.25%.
What is the amount left?The amount of substance left is calculated as follows;
N = N₀(1/2)^(t/T)
where;
N₀ is the initial amount of the substanceN is the amount remaining after time tT is the half-life of the substance,
we have;
N₀ = 100g,
T = 5670 years, and
t = 22680 years
N = 100 x (1/2)^(22680/5670)
N = 6.25 g
The fraction remaining is calculated as follows
fraction remaining = N/N₀
fraction remaining = 6.25/100
fraction remaining = 0.0625 or 6.25%
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A metal wire of diameter 1. 00 mm can support a tension of 0. 240 kn. A construction worker needs a cable made of these wires to support a tension of 20. 0 kn. The cable should have diameter (in cm) of what order of magnitude?
The cable should have a diameter of approximately 0.092 cm or an order of magnitude of one centimeter.
To determine the diameter of the cable needed to support a tension of 20.0 kN, we can use the principle of cross-sectional area. The maximum tension that a wire can withstand is proportional to its cross-sectional area. Therefore, to support a tension that is 83.33 times greater than the maximum tension of a single wire, the cross-sectional area of the cable must also be 83.33 times greater.
The cross-sectional area of a wire is given by the formula A = πr², where A is the cross-sectional area, and r is the radius of the wire. Since the diameter of the wire is given as 1.00 mm, the radius is 0.50 mm or 0.005 cm. Therefore, the cross-sectional area of a single wire is:
A₁ = π(0.005 cm)² = 0.00007854 cm²
To find the diameter of the cable, we can use the formula for the cross-sectional area of a circle:
A₂ = πr₂²
where A₂ is the cross-sectional area of the cable and r₂ is the radius of the cable.
We know that the cross-sectional area of the cable needs to be 83.33 times greater than the cross-sectional area of a single wire:
A₂ = 83.33 A₁ = 83.33 x 0.00007854 cm² = 0.00654 cm²
Substituting this value into the formula for the cross-sectional area of a circle:
πr₂² = 0.00654 cm²
r₂² = 0.00654/π
r₂ = √(0.00654/π) = 0.046 cm
Therefore, the radius of the cable is 0.046 cm, and the diameter is twice that:
d = 2r₂ = 0.092 cm or 0.92 mm (to two significant figures)
In conclusion, the cable should have a diameter of the order of magnitude of one centimeter (0.092 cm).
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1. During a summer storm, a bolt of lightning is seen. A short time later, thunder is heard. If the lightning struck 3. 50 km away, what was the time period between the lightning and thunder? The speed of sound in air is 331. 0 m/s at 0. 00 °C but the temperature is actually a warm 30. 0 °C. Show your work!
2. The following measurements were made using a Kundt’s tube generator as was done in our virtual lab. Distance from node (crest) to node (trough) = 56. 5 cm at a frequency of 894Hz. What was the velocity of sound in the tube? Knowing that the standard velocity of Helium is 1007 m/s, Air is 340 m/s and Carbon dioxide is 267 m/s, which gas was in the tube? (Assume all were at the same temperature)
The time period between the lightning and thunder is 10.09 seconds.
The velocity of sound in the tube was 1009.2 m/s
The time period between the lightning and thunder can be calculated using the equation: distance = speed × time. Since we know the distance (3.50 km) and the speed of sound at 30.0 °C (347.2 m/s), we can rearrange the equation to solve for time: time = distance / speed. Plugging in the numbers, we get: time = 3.50 km / 347.2 m/s = 10.09 seconds.
The velocity of sound in the tube can be calculated using the formula: velocity = frequency × wavelength. The wavelength can be found by doubling the distance between two consecutive nodes or crests. In this case, the wavelength is 2 × 56.5 cm = 113 cm = 1.13 m. Plugging in the frequency (894 Hz) and the wavelength (1.13 m), we get: velocity = 894 Hz × 1.13 m = 1009.2 m/s. Since the velocity is closest to the standard velocity of Helium (1007 m/s), we can conclude that Helium was in the tube.
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Which therapy is associated with light waves, but not sound waves? breaking down kidney stones acoustically targeting the delivery of a drug cauterizing an incision or wound ablating tumors
Cauterizing an incision or wound therapy is the therapy that is associated with light waves, but not sound waves. The correct option is (C).
A medical treatment called cauterizing an incision or wound includes burning or coagulating tissues with heat or electricity in order to stop bleeding or hasten wound healing. The main objective of cauterization is to produce a thermal action that closes off blood vessels in order to provide hemostasis and stop excessive bleeding.
During surgical procedures, cauterization is frequently performed to stop bleeding, remove or destroy aberrant tissue, or close off blood arteries. In some medical treatments, such as the removal of skin tags or warts, it is also utilized.
Hence, the therapy is associated with light waves, but not sound waves cauterizing an incision or wound. Option (C) is correct.
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The complete question is:
A: breaking down kidney stones
B: acoustically targeting the delivery of a drug
C: cauterizing an incision or wound
D: ablating tumors
Can someone please help me with this lesson outline?
Answer:
The amount of gravitational force INCREASES as the distance between two objects increases; thus, an astronauts weight DECREASES as she or he moves away from earth into space.
hope this helped.
1) Calculate the centripetal force acting on a 925 kg car as it rounds an unbanked curve with a radius of 75 m at a speed of 22 m/s.
2) A car with a mass of 833 kg rounds an unbanked curve in the road at a speed of 28. 0 m/s. If the radius of the curve is 105 m, what is the average centripetal force exerted on the car?
3) An amusement park ride has a radius of 2. 8 m. If the time of one revolution of a rider is 0. 98 s, what is the speed of the rider?
4) An electron (m=9. 11x10 -31kg) moves in a circle whose radius is 2. 00 x 10 -2m. If the force acting on the electron is 4. 60x10 -14N, what is its speed?
5) A 2. 7x10 3kg satellite orbits the Earth at a distance of 1. 8x10 7m from the Earth’s centre at a speed of 4. 7x10 3m/s. What force does the Earth exert on the satellite?
6) A string can withstand a force of 135 N before breaking. A 2. 0 kg mass is tied to the string and whirled in a horizontal circle with a radius of 1. 10 m. What is the maximum speed that the mass can be whirled at before the string breaks?
7) A motocross rider at the peak of his jump has a speed such that his centripetal acceleration is equal to g. As a result, he does not feel any supporting force from the seat of his bike, which is also accelerating at rate g. Therefore, he feels if there is ni force of gravity on him, a condition described as apparent weightlessness. If the radius of the approximately circular jump is 75. 0 m, what is the speed of the bike?
The centripetal force is 5,444.27 N, the average centripetal force exerted on a car is 6,988.31 N, the speed of the rider is 18.06 m/s, the speed of an electron is 1.73 x 10⁷ m/s, the force exerted by the Earth on a satellite is 1.84 x 10⁴ N, the maximum speed is 27.39 m/s and the speed of the bike is 27.39 m/s.
1. The centripetal force acting on a 925 kg car as it rounds an unbanked curve with a radius of 75 m at a speed of 22 m/s can be calculated using the formula [tex]Fc = (mv^{2} )/r[/tex]. Substituting the given values, we get [tex]Fc = (925 kg \times 22^{2} m^{2} / s^{2} ) / 75m[/tex] = 5,444.27 N.
2. To find the average centripetal force exerted on a car with a mass of 833 kg rounding an unbanked curve with a radius of 105 m at a speed of 28.0 m/s, we can use the same formula [tex]Fc = (mv^{2} )/r[/tex]. Substituting the given values, we get [tex]Fc = (833 kg \times 28.0^{2} m^{2} /s^{2} ) / 105 m[/tex] = 6,988.31 N.
3. The speed of the rider in an amusement park ride with a radius of 2.8 m and a time of one revolution of 0.98 s can be calculated using the formula [tex]v = 2\pi r / t[/tex]. Substituting the given values, we get[tex]v = (2 \times 3.14 \times 2.8 m) / 0.98 s[/tex] = 18.06 m/s.
4. The speed of an electron in a circle with a radius of [tex]2.00 \times 10^{-2} m[/tex] and a force [tex]4.60 \times 10^{-14} N[/tex] acting on it can be calculated using the formula [tex]v = \sqrt{(Fcr / m)}[/tex]. Substituting the given values, we get
[tex]v = \sqrt{[(4.60 \times 10^{-14} N \times 2.00 x 10^{-2} m) / 9.11 \times 10^{-31} kg]}[/tex]
[tex]= 1.73 \times 10^7 m/s.[/tex]
5. The force exerted by the Earth on a satellite with a mass of [tex]2.7 \times 10^3[/tex] kg orbiting at a distance of [tex]1.8 \times 10^7[/tex] m and a speed of [tex]4.7 \times 10^3\;m/s[/tex] can be calculated using the formula [tex]Fg = (Gm_{1} m_{2}) / r^{2}[/tex]. Substituting the given values, we get
[tex]Fg = (6.67 \times 10^{-11} N(m/kg)^2 \times 5.97 \times 10^{24} kg \times 2.7 \times 10^3 kg) / (1.8 \times 10^7 m)^{2}[/tex]
[tex]= 1.84 \times 10^4 N.[/tex]
6. The maximum speed at that a 2.0 kg mass can be whirled in a horizontal circle with a radius of 1.10 m before the string breaks, given a maximum force of 135 N that the string can withstand, can be calculated using the formula[tex]v = \sqrt(Fr / m)[/tex]. Substituting the given values, we get
[tex]v = \sqrt{[(135 N \times 1.10 m) / 2.0 kg]}[/tex]
= 16.47 m/s.
7. The speed of the bike in a motocross jump with a radius of 75.0 m, where the rider experiences apparent weightlessness due to the acceleration of the bike, can be calculated using the formula [tex]v = \sqrt{(rg)[/tex]. Substituting the given values, we get
[tex]v = \sqrt{(75.0\;m \times 9.81 m/s^{2} )}[/tex]
= 27.39 m/s.
In summary, these problems involve calculating various aspects of circular motion, including centripetal force, speed, and radius, using different formulas. The calculations involve substituting the
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An electronic device requires a power of 15 w when connected to a 9.0-v battery. how much power is delivered to the device if it is connected to a 6.0-v battery
The power delivered to the device when connected to a 6.0 V battery is 10 W, which is less than the power delivered when connected to a 9.0 V battery.
The power delivered to the electronic device is proportional to the voltage supplied to it.
The relationship between power, voltage, and current is given by the equation P = VI, where P is power, V is voltage, and I is current. In this case, the power is given as 15 W when the device is connected to a 9.0 V battery.
Using the equation P = VI, we can solve for the current as I = P/V = 15 W / 9.0 V = 1.67 A. When the device is connected to a 6.0 V battery, the power delivered to the device can be calculated as P = VI = 1.67 A x 6.0 V = 10 W.
Therefore, the power delivered to the device when connected to a 6.0 V battery is 10 W, which is less than the power delivered when connected to a 9.0 V battery.
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When energy leaves the sun’s core, it travels through the radiative zone in the form of ___.
a. liquids
b. coronas
c. electromagnetic waves
d. convection currents
When energy leaves the sun's core, it travels through the radiative zone in the form of c. electromagnetic waves.
This is also called as radiative energy. The radiative zone is the second zone of the sun, and it is the region where the energy created by nuclear reactions in the core is transferred through the Sun's outer layers. In this zone, the energy moves in the form of photons, which are particles of light that carry the energy.
The radiative zone is characterized by the high temperature and density of its materials, which cause the photons to scatter frequently before they can escape the zone. The photons that make it through the radiative zone eventually reach the convective zone, where they transfer their energy to the gas particles that rise and fall in the Sun's atmosphere through convection currents. These currents help distribute the energy from the core to the outer layers of the Sun and eventually to space.
In summary, the correct answer to the question is c. electromagnetic waves, which travel through the radiative zone as particles of light or photons.
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Why are Buildings in UAE made with glazed glass?
The buildings in UAE made with glazed glass is because of their low maintenance and ease of installation.
Glazed glass means the glass that is used for buildings and architectural purposes. The glass facades are specifically energy efficient and support green glazing. In UAE buildings, the majority of glazing is to achieve a shading co-efficient of 0.25 which results in high-performance glazing. Glass transmits up to 80% of natural daylight and ensures cost savings.
Glass buildings help in energy efficiency in Eastern countries. Adopting glass on buildings is not easy in Middle Eastern countries due to extreme weather conditions. But the technologies made it possible and glass become a widely used material in the Middle East.
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A typical color television draws about 2. 5 A
when connected to an 89 V source.
What is the effective resistance of the T. V.
set?
Answer in units of Ω
The effective resistance of the TV set is 35.6 ohms (Ω).
To find the effective resistance of the TV set, we can use Ohm's Law, which states that Voltage (V) = Current (I) × Resistance (R). We need to rearrange the formula to solve for resistance: R = V / I.
Given the information in your question:
Current (I) = 2.5 A
Voltage (V) = 89 V
Now we can calculate the resistance (R):
R = V / I
R = 89 V / 2.5 A
R = 35.6 Ω
The effective resistance of the TV set is 35.6 ohms (Ω).
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2. A woman lifts up a laundry basket 1. 5m and carries it 20m across the room. This takes 15s.
Work is done on the laundry basket_*
(20 Points)
in walking across the room
during the entire 15s
work is not done
in lifting the basket
A woman lifts up a laundry basket 1.5m and carries it 20m across the room in 15s.
Work is done on the laundry basket both in lifting the basket and in walking across the room during the entire 15s.
Lifting the basket:
The work done in lifting the basket is equal to the force applied multiplied by the distance lifted. The force applied is the weight of the basket, which is equal to the mass of the basket multiplied by the acceleration due to gravity (9.8 m/s^2).
Let's assume the mass of the basket is 'm'. The work done in lifting the basket vertically is given by:
Work_lift = force_lift × distance_lift = (m × 9.8) × 1.5
Carrying the basket horizontally:
When the woman carries the basket across the room, work is done against the force of friction between the basket and the floor. The work done is equal to the force of friction multiplied by the distance traveled horizontally.
The force of friction can be calculated using the coefficient of friction (μ) and the normal force (N). The normal force is equal to the weight of the basket since it is on a horizontal surface.
Let's assume the coefficient of friction between the basket and the floor is 'μ'. The work done in moving the basket horizontally is given by:
Work_horizontal = force_friction × distance_horizontal = (μ × m × 9.8) × 20
The total work done on the basket during the entire 15s is the sum of the work done in lifting and the work done horizontally:
Total work = Work_lift + Work_horizontal
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One who is capable of identifying existing and predictable.
It seems like the phrase you provided is incomplete or ambiguous. However, based on the partial phrase you provided, "One who is capable of identifying existing and predictable," it could refer to a person who has the ability to recognize and understand things that currently exist and can be predicted in the future.
This could describe someone who has a strong analytical or observational skills and can perceive patterns, trends, or regularities in various aspects of life, such as in scientific phenomena, financial markets, human behavior, or other areas where predictability and existing patterns are sought.
If you have a specific context or a more detailed question, please provide additional information, and I'll be glad to provide a more specific response.
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When you pedal really fast on a bike, you can feel the wind slowing you down.
Which force causes this?
O
A. Strong nuclear force
B. Magnetic force
O
OD. Gravity
C. Air resistance
Answer:
the answer is the option C
A 50. 0-kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 250 n at an angle of 32. 0° above the horizontal. The coefficient of kinetic friction between the box and the surface is 0. 350. What is the acceleration of the box?.
The acceleration of the box can be determined using Newton's second law of motion, where the net force acting on the box is equal to the mass of the box multiplied by its acceleration.
In this case, the net force acting on the box is equal to the force of the rope (250 n at an angle of 32.0° above the horizontal) minus the force of kinetic friction (0.350 × 50 kg × 9.81 m/s2). After solving for the acceleration, the acceleration of the box is 5.3 m/s2.
To summarise, the acceleration of a box being pulled along a horizontal surface with a force of 250 n at an angle of 32.0° above the horizontal and a coefficient of kinetic friction of 0.350 is 5.3 m/s2. This acceleration can be determined by using Newton's second law of motion and calculating the net force acting on the box.
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Two wind turbines are set up with the following conditions: Turbine Ahas blades that are twice as long as the blades on Turbine B. The tips of the blades on Turbine A are moving twice as fast as the tips of the blades on Turbine B. Part D Which turbine takes the lesser amount of time to rotate through 1.0 radian of angular displacement? A. turbine A a turbine B They take the same amount of time. The answer cannot be determined from the information given. ; Subrnit Request Answer Part 5 29 As in Part D, two wind turbines with different length blades are rotating. Consider what needs to happen in order to change the angular speed of one of the turbines. If the turbine is to spin more quickly, should the angular acceleration, a be positive or negative?B. a should be positive.C. a should be negative. D. We cannot tell which direction a should be without knowing the direction of the angular velocity,
If a wind turbine with different length blades needs to spin more quickly, the angular acceleration should be positive. The correct answer is B
Part A: In the given scenario, Turbine A has blades twice as long as Turbine B, and the tips of the blades on Turbine A are moving twice as fast as the tips of the blades on Turbine B. Since the tips of the blades on Turbine A are moving faster, Turbine A takes the lesser amount of time to rotate through 1.0 radian of angular displacement. So, the correct answer is (a) Turbine A.
Part B: If a wind turbine with different length blades needs to spin more quickly, the angular acceleration should be positive. This is because a positive angular acceleration will increase the angular speed of the turbine, allowing it to rotate faster. So, the correct answer is (b) should be positive.
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Complete question:
Two wind turbines are set up with the following conditions: Turbine A has blades that are twice as long as the blades on Turbine B. The tips of the blades on Turbine A are moving twice as fast as the tips of the blades on Turbine B.
Part A. Which turbine takes the lesser amount of time to rotate through 1.0 radian of angular displacement?
a. turbine A
b. They take the same amount of time.
c. turbine B
d. The answer cannot be determined from the information given.
Part B. As in Part A, two wind turbines with different length blades are rotating. Consider what needs to happen in order to change the angular speed of one of the turbines. If the turbine is to spin more quickly, should the angular acceleration, be positive or negative?
a. should be negative
b. should be positive
c. We cannot tell which direction it should be without knowing the direction of the angular velocity
Two objects of the same mass travel in opposite directions along a horizontal surface. Object x has a speed of 5ms and object y has a speed of 5ms, as shown in the figure. After a period of time, object x collides with object y. In scenario 1, the objects stick together after the collision. In scenario 2, the objects do not stick together after the collision
After the collision, the two objects will stick together and move with a velocity of 5ms in the same direction.
What is collision?Collision is a physical phenomenon that occurs when two or more objects interact with enough force to cause damage to one or more of the objects. This can occur when two objects come into contact with each other, or when two objects are moving at different speeds and collide with each other. Collisions can be caused by a variety of factors, including the speed and mass of the objects, the angle of their contact, and the surface area of the objects.
Scenario 1:
After the collision, the two objects will stick together and move with a velocity of 5ms in the same direction.
Scenario 2:
After the collision, the two objects will bounce off each other and move in opposite directions with the same velocity of 5ms.
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Consider example 20. 15, what angle of deflection would you get if the electron gun distance as well as electron defelctor distance were to both double, with the electric fields staying as in the example?.
If the electron gun distance and electron deflection distance both double, while the electric fields stay the same, then the angle of deflection would also double.
This is because the electric field strength is directly proportional to the angle of deflection, and since the electric field strength is staying the same, the angle of deflection increases proportionally with the increase in distance.
The equation to determine the angle of deflection is as follows: θ = Vd/E, where θ is the angle of deflection, V is the velocity of the electron, d is the distance between the electron gun and deflection plate, and E is the strength of the electric field.
When the distance between the two plates doubles, the angle of deflection will also double. Therefore, if the electron gun and electron deflection plate are both doubled in distance, the angle of deflection would be double the original angle.
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A wood block of mass m rests on a larger wood block of mass M that rests on a wooden table. The coefficients of static and kinetic friction between all surfaces are μs and μk , respectively.
A)What is the minimum horizontal force, F , applied to the lower block that will cause it to slide out from under the upper block? To solve this problem, assume that the force is applied so suddenly that both blocks slip at the same time. Express your answer in terms of some or all of the variables m , M , μs , μk , and appropriate constants. F=?
The minimum horizontal force required to cause the lower block to slide out from under the upper block is F = μs(Mg + mg)
How to calculate the forceLet's consider the forces acting on the lower block. The weight of the block is mg, where g is the acceleration due to gravity. The normal force acting on the block is N = Mg + mg, where M is the mass of the upper block. The maximum static frictional force that can act between the two blocks is μsN.
If the applied force is F, the net force acting on the lower block is F - μsN. If this net force is greater than zero, the block will slide. Therefore, we can write:
F - μsN > 0
Substituting for N, we get:
F - μs(Mg + mg) > 0
Solving for F, we get:
F > μs(Mg + mg)
Therefore, the minimum horizontal force required to cause the lower block to slide out from under the upper block isF = μs(Mg + mg).
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You do 25 J of work in pushing a crate up a ramp. If the output work from the
inclined plane is 10 J, then what is the efficiency of the inclined plane?
The efficiency of the inclined plane is 40%.
The efficiency of the inclined plane can be calculated by dividing the output work by the input work and multiplying by 100% to get a percentage.
Efficiency = (Output work / Input work) x 100%
In this case, the input work is 25 J and the output work is 10 J.
Efficiency = (10 J / 25 J) x 100%
Efficiency = 0.4 x 100%
Efficiency = 40%
Therefore, the efficiency of the inclined plane is 40%.
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a military airplane sometimes needs to refuel in mid-air, and another plane flies above it to add the fuel. if a plane with a mass of 7,800 kg is traveling with a velocity of 30 m/s and refuels by adding an additional 800 kg of jet fuel, what will its new velocity be?
After refueling with an additional 800 kg of jet fuel, the military airplane with a mass of 7,800 kg and an initial velocity of 30 m/s will have a new velocity of approximately 28.1 m/s.
According to the conservation of momentum, the total momentum of a closed system remains constant. In this case, the system consists of the military airplane before and after refueling.
Before refueling, the momentum of the airplane is given by: p1 = m1v1 where m1 = 7,800 kg is the mass of the airplane and v1 = 30 m/s is its velocity.
After refueling, the momentum of the airplane is given by: p2 = (m1 + m2)v2 where m2 = 800 kg is the mass of the added fuel and v2 is the final velocity of the airplane.
Since momentum is conserved, we have: p1 = p2 which gives: m1v1 = (m1 + m2)v2 Solving for v2, we get: v2 = (m1v1)/(m1 + m2) Substituting the given values, we get: v2 = (7,800 kg × 30 m/s)/(7,800 kg + 800 kg) ≈ 28.1 m/s
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