Galileo used marbles rolling down inclined planes to deduce some basic properties of constant accelerated motion. In particular, he measured the distance a marble rolled during specific time periods. For example, suppose a marble starts from rest and begins rolling down an inclined plane with constant acceleration a. After 1 s, you find that it moved a distance .
a. In terms of x, how far does it move in the next 1 s time period—that is, in the time between 1 s and 2 s?
b. How far does it move in the next second of the motion?
c. How far does it move in the nth second of the motion?
Answer:
a) y₁ = ½ a, b) y₂ = 4 y₁, c) y₃ = 9 y₁
Explanation:
For this exercise we can use the accelerated motion relationships.
Let's set a reference system where the x axis is parallel to the plane and its positive side is going down the plane.
y = y₀ + v₀ t + ½ a t²
in that case where we throw the marble is the zero point, y₀ = 0, as part of rest its initial velocity is zero v₀ = 0 and a is the acceleration along the inclined plane
y = ½ a t²
a) in the first second t = 1
y₁ = ½ a
b) in the next second of movement
t = 2 s
y₂ = ½ a 2²
y₂ = 4 ½ a
y₂ = 4 y₁
c) for the next second
t = 3 s
y₃ = ½ a 3²
y₃ = 9 ½ a
y₃ = 9 y₁
This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Communication satellites are placed in a geosynchronous orbit, i.e., in a circular orbit such that they complete one full revolution about the earth in one sidereal day (23.934 h), and thus appear stationary with respect to the ground. Determine the altitude of these satellites above the surface of the earth in both SI and U.S. customary units. The altitude is SI units in km. The altitude is U.S. customary units in mi.
Answer:
the altitude of these satellites above the surface of the earth;
35790 km ( SI units )
22243.63 miles ( U.S. customary unit )
Explanation:
Given the data in the question;
Time taken by the satellite to complete on revolution is 23.934 hours
= 23.934 × 60 × 60 = 86162.4 seconds
now, let h represent altitude, r represent Orbit radius, v represent Orbit speed.
we know that
v² = GM/r
= gR²/r
= (9.81m/s² × ( 6.37 × 10⁶ m)²) / r
v = 19.95 × 10⁶ / √r ---------- let this be equation
also;
time t = 2πr/v
86162.4 s = 2πr/v -------- let this be equation 2
a) Determine the altitude of these satellites above the surface of the earth in both SI and U.S. customary units
we substitute v in equation into equation 2
so
86162.4 s = 2πr / (19.95 × 10⁶ / √r)
r = 42.16 × 10⁶ m
so, altitude h = r - R
h = 42.16 × 10⁶ m - 6.37 × 10⁶ m
h = 35790000 m
convert to kilometer
h = 35790000 / 1000
h = 35790 km
Convert to miles
h = 35790 / 1.609
h = 22243.63 miles
Therefore, the altitude of these satellites above the surface of the earth;
35790 km ( SI units )
22243.63 miles ( U.S. customary unit )
3. Sodium-24 has a half-life of 15 hours. If a sample of sodium-24 has an
original activity of 800 Bq, what will
its activity be after:
i) 15 hours?
ii) 30 hours?
iii) 45 hours?
iv) 60 hours?
Answer:
See explanation
Explanation:
From the formula;
0.693/t1/2 = 2.303/t log (Ao/A)
t1/2 = half life of Sodium-24
Ao = initial activity of Sodium-24
A= activity of Sodium-24 at time = t
So,
0.693/15 = 2.303/15 log (800/A)
0.0462 = 0.1535 log (800/A)
0.0462/0.1535 = log (800/A)
0.3 = log (800/A)
Antilog(0.3) = (800/A)
1.995 = (800/A)
A = 800/1.995
A = 401 Bq
ii) 0.693/15 = 2.303/30 log (800/A)
0.0462 = 0.0768 log (800/A)
0.0462/0.0768 = log (800/A)
0.6 = log (800/A)
Antilog (0.6) = (800/A)
3.98 = (800/A)
A = 800/3.98
A = 201 Bq
iii)
0.693/15 = 2.303/45 log (800/A)
0.0462 = 0.0512 log (800/A)
0.0462/0.0512 = log (800/A)
0.9 = log (800/A)
Antilog (0.9) = (800/A)
7.94 = (800/A)
A = 800/7.94
A= 100.8 Bq
iv)
0.693/15 = 2.303/60 log (800/A)
0.0462 = 0.038 log (800/A)
0.0462/0.038 = log (800/A)
1.216 = log (800/A)
Antilog(1.216) = (800/A)
16.44 = (800/A)
A = 800/16.44
A = 48.66 Bq
current must flow if 0.56 coulombs is to be transferred 35ms
Answer:
the current is 16 amphere
Explanation:
The computation of Current is shown below:
As we know that
1 ms = 0.001s
So for 35 ms = 0.035
Now the current is
= 0.56 ÷ 0.035
= 16 AMphere
Hence, the current is 16 amphere
Calculate the magnitude of the gravitational force exerted by Mercury on a 70 kg human standing on the surface of Mercury. (The mass of Mercury is 3.31023 kg and its radius is 2.4106 m.)
Answer:
2.66×10⁻⁹ N.
Explanation:
From the question,
Applying newton's law of universal gravitation,
Fg = GMm/r²............................... Equation 1
Where Fg = gravitational force, G = universal constant, M = mass of the mercury, m = mass of the human, r = radius of Mercury
Given: M = 3.31023 kg, M = 70 kg, r = 2.4106
Constant: G = 6.67×10⁻¹¹ Nm²/kg²
Substitute these values into equation 1
Fg = 6.67×10⁻¹¹(70×3.31023)/(2.4106²)
Fg = 2.66×10⁻⁹ N.
A baseball is thrown horizontally from a cliff at 30 m/s and lands 7 seconds after the baseball was thrown. Calculate the horizontal AND vertical distance.
Answer:
The horizontal and vertical distances are x = 210 m and y = -240.35 m, respectively.
Explanation:
Using the equation of the displacement in the x-direction, we have:
(let's recall we have a constant velocity in this direction)
[tex]x=v_{ix}t[/tex]
Where:
v(ix) is the initil velocity in the x direction (v(ix) = 30 m/s)t is the time (t = 7 s)[tex]x=30(7)[/tex]
[tex]x=210\: m[/tex]
Now, we need to use the equation of the displacement in the y-direction to find the vertical distance. Here we have an acceleration (g)
[tex]y=v_{iy}t-\frac{1}{2}gt^2[/tex]
Where:
v(iy) is the initial velocity at the y-direction. In this case, it will be 0t is the timeg is the acceleration of gravity (g=9.81 m/s²)Then, the vertical position at 7 s is:
[tex]y=-\frac{1}{2}(9.81)(7)^2[/tex]
[tex]y=-240.35\: m[/tex]
Therefore, the horizontal and vertical distances are x = 210 m and y = -240.35 m, respectively. The minus sign means the negative value in the y-direction.
I hope it helps you!
because there the sperm and eggs are combining together to produce
so thats why they look alike
Answer:
yes. that is how a baby is conceived.
2. A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling at 750 m/s. How much work was done by the rocket? What is the magnitude of the acceleration of the rocket? And how long did the flight take?
Answer:
797700000 J
Explanation:
From the question,
The work done by the rocket, is given as,
W = Ek+Ep............. Equation 1
Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.
Ep = mgh............ Equation 2
Ek = 1/2mv²............. equation 3
Substitute equation 2 and equation 3 into equation 1
W = mgh+1/2mv².............. Equation 4
Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.
Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²
Substitute into equation 4
W = 2000(12000)(9.8)+1/2(2000)(750²)
W = 235200000+562500000
W = 797700000 J
Which statement is correct?
A. If the electric field is zero everywhere inside a region of space, the potential must also be zero in that region.
B. When the electric field is zero at a point, the potential must also be zero there.
C. If the electrical potential in a region is constant, the electric field must be zero everywhere in that region.
D. If the electric potential at a point in space is zero, then the electric field at that point must also be zero.
Answer:
The answer is "Choice C ".
Explanation:
The relationship between the E and V can be defined as follows:
[tex]\to E= -\Delta V[/tex]
Let,
[tex]\to E= \frac{\delta V}{\delta x}[/tex]
When E=0
[tex]\to \frac{\delta V}{\delta x}=0[/tex]
v is a constant value
Therefore, In the electric potential in a region is a constant value then the electric-field must be into zero that is everywhere in the given region, that's why in this question the "choice c" is correct.
one car travels due east at 40 km/hr and a second car travels north at 40km/hr. Are their velocities equal?
Answer:
No.
Explanation:
Velocity is a vector quantity which means that it has a certain direction so things that move in different directions DO NOT have the same velocity.
Which statement best compares potential and kinetic energy?
O Objects always have more potentiał energy than kinetic energy.
O Kinetic energy increases and potential energy decreases when the velocity of an object increases
O Only potential energy decreases when an object's height increases.
O Objects always have more kinetic energy than potential energy.
Answer:
Kinetic energy increases and potential energy decrease when velocity of an object increase.
If the diameter of a moose eye is 40 mm, what is the total refractive power of the anterior portion of the eye?
Answer:
-the ratio of the speed of light
in air to the speed of light in the substance.
-speed of light in air 300,000 km/sec, which decreases when it passes through a transparent substance.
-e.g.. speed of light in substance = 200,000 km/sec, R.I. = 300,000/200,000 = 1.5
Explanation:
Astronauts need sophisticated spacesuits to protect them from the harsh conditions of space. These spacesuits are very heavy for astronauts to wear on Earth. For an astronaut on the Moon, however, the suit would seem lighter. Which of the following statements explains why the spacesuit would feel lighter on the Moon?
A. There is no gravity on the Moon.
B. Earth’s gravity can’t be felt from the Moon.
C. The force of the Moon’s gravity is less than Earth’s.
D. The force of the Sun’s gravitational attraction varies throughout the Solar System and is stronger closer to Earth.
Answer:
I think it’s c
Explanation:
A solar panel is used to collect energy from the sun and change it into other forms of energy. The picture below shows some solar panels on the roof of a building. Which form of energy to collected by the solar panels?
A. Wind
B. sound
C. Magnetic
D. Light
A dog finds a toy at rest on the floor. The dog pushes the toy horizontally on a frictionless floor with a net force of 2.0 Newtons for 3.0 meters. How much kinetic energy does the toy gain? Round your answer to the nearest whole number.
Answer:
the kinetic energy gained by the toy is 6J.
Explanation:
Given;
net applied to the toy by dog, F = 2 N
distance moved by the toy, d = 3 m
Apply the principle of work-energy theorem to determine the kinetic energy gained by the toy.
ΔK.E = W
= F x d
= 2 x 3
= 6 J
Therefore, the kinetic energy gained by the toy is 6J.
A harmonic oscillator has mass 0.500 kg and an ideal spring with force constant 140 N/m. Find:
a. the period
b. the frequency
c. the angular frequency
Answer:
T = 0.375 s, f = 2.66 Hz and ω = 16.71 rad/s
Explanation:
Given that,
The mass of a harmonic oscillator, m = 0.5 kg
The force constant of the spring, k = 140 N/m
The frequency of a harmonic oscillator is given by :
[tex]f=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}[/tex]
Substitute all the values,
[tex]f=\dfrac{1}{2\pi }\sqrt{\dfrac{140}{0.5}} \\\\f=2.66\ Hz[/tex]
Time period is given by :
[tex]T=\dfrac{1}{f}\\\\T=\dfrac{1}{2.66}\\\\T=0.375\ s[/tex]
The angular frequency is given by :
[tex]\omega=2\pi f\\\\\omega=2\pi \times 2.66\\\\\omega=16.71\ rad/s[/tex]
Hence, this is the required solution.
1. Suppose the spring in Sample Problem A is replaced with a spring that stretches
36 cm from its equilibrium position.
a. What is the spring constant in this case?
b. Is this spring stiffer or less stiff than the one in Sample Problem A?
Answer:
a. spring constant = 125 N/m
b. This spring is less stiff than the one in Sample Problem A.
Explanation:
P.S - The Sample Problem A is as follows :
Given - Sample Problem A - A load of 45 N attached to a spring that is
hanging vertically stretches the spring 0.14 m. What is the spring
constant?
Suppose the spring in Sample Problem A is replaced with a spring
that stretches 36 cm from its equilibrium position.
To find - a. What is the spring constant in this case?
b. Is this spring stiffer or less stiff than the one in Sample Problem A.
Proof -
As given,
Load = 45 N
Amplitude = 0.14 m
Let the spring constant = k
As we know that,
Load = k (Amplitude)
⇒45 = k(0.14)
⇒k = [tex]\frac{45}{0.14}[/tex] = 321.43
∴ we get
Spring constant in Sample problem A = 321.43
Now,
a.)
Given, Amplitude = 36 cm = 0.36 m
Let the spring constant = k₁
⇒45 = k₁ (0.36)
⇒k₁ = [tex]\frac{45}{0.36}[/tex] = 125 N/m
b.)
AS we can see that k₁ < k
⇒ This spring is less stiff than the one in Sample Problem A.
a. spring constant = 125 N/m
b. This spring is less stiff than the one in Sample Problem A.
what is spring constant?The spring constant generally shows the stiffness of the spring and is the ratio of the force applied to the deflection of the spring.
It is given in the question that:
Sample Problem A - A load of 45 N attached to a spring that is
hanging vertically stretches the spring 0.14 m.
Suppose the spring in Sample Problem A is replaced with a spring
that stretches 36 cm from its equilibrium position.
a. What is the spring constant in this case?
As given,
Load F = 45 N
Amplitude x= 0.14 m
Let the spring constant = k
As we know that spring force will be
[tex]F=k\times x[/tex]
[tex]45=k\times 0.14[/tex]
⇒k = 321.43N/m
∴ we get
Spring constant in Sample problem A is [tex]k=321.43\ \frac{N}{m}[/tex]
Now,
Given, Amplitude = 36 cm = 0.36 m
Let the spring constant = k₁
⇒45 = k₁ (0.36)
⇒k₁= 125 N/m
b.) Is this spring stiffer or less stiff than the one in Sample Problem A.
AS we can see that k₁ < k new spring is less stiff than the one in Sample Problem A.
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Imagine that 10.0 g of liquid helium, initially at 4.20 K, evaporate into an empty balloon that is kept at 1.00 atm pressure. What is the volume of the balloon at (a) 25.0 K and (b) 293 K?
Answer:
(a) The volume of the liquid helium at 25 K is 5.13 L
(b) The volume of the liquid helium at 293 K is 60.14 L.
Explanation:
Given;
mass of the liquid helium, m = 10 g
initial temperature of the liquid helium, T₁ = 4.2 K
pressure of the liquid helium, P = 1.00 atm
Atomic mass of Helium, = 4 g
number of moles of Helium, n = 10 / 4 = 2.5 moles
The initial volume of the liquid helium is calculated as;
[tex]PV_1 = nRT_1\\\\V_1 = \frac{nRT_1}{P} \\\\[/tex]
where;
R is ideal gas constant, = 0.08205 L.atm./mol.K
[tex]V_1 = \frac{2.5 \times 0.08205 \times 4.2}{1 } \\\\V_1 = 0.862 \ L[/tex]
(a) The volume of the liquid helium at 25 K.
Apply Charles law;
[tex]\frac{V_1}{T_1} =\frac{V_2}{T_2} \\\\V_2 = \frac{V_1T_2}{T_1} \\\\V_2 = \frac{0.862 \times 25 }{4.2} \\\\V_2 = 5.13 \ L[/tex]
(b) The volume of the liquid helium at 293 K.
[tex]\frac{V_1}{T_1} =\frac{V_2}{T_2} \\\\V_2 = \frac{V_1T_2}{T_1} \\\\V_2 = \frac{0.862 \times 293 }{4.2} \\\\V_2 = 60.14 \ L[/tex]
A diesel engine lifts the 225 kg hammer of a pile driver 20 m in 5 seconds. How much work is done on
the hammer? What is the power?
Answer:
a. Workdone = 44100 Joules
b. Power = 8820 Watts.
Explanation:
Given the following data:
Mass = 225kg
Distance = 20m
Time = 5 seconds
To find the workdone;
Workdone = force * distance
But force = mg
We know that acceleration due to gravity is equal to 9.8m/s²
Force = 225*9.8 = 2205N
Substituting the values into the equation, we have;
Workdone = 2205 * 20
Workdone = 44100 Joules
b. To find the power;
Power = workdone/time
Power = 44100/5
Power = 8820 Watts.
Which cell part controls all the other parts of a cell?
Answer:
Nucleus
Explanation:
Answer: Nucleolus
Explanation: The nucleolus is like the cells brain. It controls all the other organelles (cell parts).
Hope I helped!
The coefficient of kinetic friction between the tires of a car and a horizontal road surface is 0.52. If the car is traveling at an initial speed of 25 m/s, and then slams on the breaks so the car skids straight ahead to a stop, how far does the car skid?
Answer:
The car skids in a distance of 61.275 meters.
Explanation:
Since the only force exerted on the car is the kinetic friction between the car and the horizontal road, deceleration of the vehicle ([tex]a[/tex]), measured in meters per square second, is determined by the following expression:
[tex]a = \mu_{k}\cdot g[/tex] (1)
Where:
[tex]\mu_{k}[/tex] - Coefficient of kinetic friction, no unit.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]\mu_{k} = 0.52[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], then the net deceleration of the vehicle is:
[tex]a = 0.52\cdot \left(-9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]a = -5.1\,\frac{m}{s^{2}}[/tex]
The distance covered by the car is finally calculated by this kinematic expression:
[tex]\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex] (2)
Where:
[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speed, measured in meters per second.
[tex]a[/tex] - Net deceleration, measured in meters per square second.
If we know that [tex]v_{o} = 25\,\frac{m}{s}[/tex], [tex]v = 0\,\frac{m}{s}[/tex] and [tex]a = -5.1\,\frac{m}{s^{2}}[/tex], then the distance covered by the car is:
[tex]\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(25\,\frac{m}{s} \right)^{2}}{2\cdot \left(-5.1\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\Delta s = 61.275\,m[/tex]
The car skids in a distance of 61.275 meters.
1. Weather factors include
a. average air temperature.
b. annual precipitation.
c. humidity.
d. two of the above
2. The dew point is the temperature at which
a. dew forms on surfaces.
b. water vapor starts to condense.
c. relative humidity is 100 percent.
d. all of the above
3. Relative humidity may decrease if
a. water vapor condenses out of the air.
b. water evaporates into the air.
c. air temperature decreases.
d. two of the above
4. Which type of cloud forms at high altitudes?
a. cirrocumulus
b. altocumulus
c. stratocumulus
d. nimbostratus
5. Which type of cloud forms when strong air currents carry warm air upward?
a. cirrus
b. stratus
c. cumulus
d. cirrostratus
6. The type of fog that forms
6. The type of fog that forms when cool air moves over a warm lake is called
a. radiation fog.
b. advection fog.
c. steam fog.
d. upslope fog.
7. Rain that passes through a layer of freezing air near the ground become
a. glaze.
b. hail.
c. sleet.
d. snow.
Answer:
1. D
Climate is generally defined as the weather condition that prevails in a particular region over a long period of time. Climate is usually measured by examining the pattern of variation in several climatic factors such as rainfall, temperature, relative humidity, wind, pressure, etc. While the weather of a place can change within a space of few hours, it takes years for a change in climatic condition to occur.
2. d
3. c
4.a.
5. c
6. a.
7. c
Explanation:
The correct answers are (1) d. two of the above (average air temperature and humidity), (2)c. the relative humidity is 100 percent, (3)d. two of the above (water vapor condenses out of the air and air temperature decreases), (4)a. cirrocumulus, (5)c. cumulus, (6)c. steam fog, and (7)c. sleet.
What is temperature?Temperature is a physical quantity that measures the average kinetic energy of the particles in a substance or system. It is a measure of how hot or cold something is, and is typically measured in units such as degrees Celsius or Fahrenheit. Temperature can also be thought of as a measure of the direction in which heat energy flows, with heat energy naturally flowing from areas of higher temperature to areas of lower temperature.
Here in the Question,
1. Weather factors include d. two of the above (average air temperature and annual precipitation are two factors that affect weather, but humidity is also an important factor that can influence the feel of the air).
2. The dew point is the temperature at which b. water vapor starts to condense. When air cools, it can reach a point where it is unable to hold all of its moisture in the form of water vapor. At this point, the water vapor starts to condense into visible droplets, such as dew, and the temperature at which this happens is called the dew point. When the dew point is reached, the relative humidity is at 100 percent.
3. Relative humidity may decrease if d. two of the above (water vapor condenses out of the air and air temperature decreases) occur. If the air cools and reaches the dew point, water vapor will start to condense into droplets, which can reduce the amount of water vapor in the air and lower the relative humidity. Similarly, if the temperature drops without any change in water vapor content, the relative humidity will decrease because colder air can hold less moisture than warmer air.
4. The type of cloud that forms at high altitudes is a. cirrocumulus. These clouds are typically found at altitudes above 18,000 feet and are characterized by small, white, puffy clouds arranged in rows or ripples. They are often a sign of fair weather but can also indicate an approaching storm.
5. The type of cloud that forms when strong air currents carry warm air upward is c. cumulus. Cumulus clouds are large, puffy clouds that can develop vertically, forming a towering cloud with a flat top. They are often associated with thunderstorms and can produce heavy rain, hail, and lightning.
6. The type of fog that forms when cool air moves over a warm lake is c. steam fog. Steam fog, also called evaporation fog or sea smoke, occurs when cold, dry air moves over a warm, moist surface and causes water vapor to rise and condense into fog. This type of fog is often seen over bodies of water during the fall and winter.
7. Rain that passes through a layer of freezing air near the ground becomes c. sleet. Sleet is formed when raindrops fall through a layer of freezing air near the ground and freeze into small ice pellets before hitting the surface. It is different from hail, which forms in strong thunderstorms when updrafts carry raindrops upward into colder air where they freeze and then fall back to the ground, and snow, which forms in clouds when water vapor freezes directly into ice crystals. Glaze is a type of ice that forms when rain falls onto a surface that is below freezing, forming a layer of ice on top of the surface.
Therefore, The correct answers are:1. Weather factors include average air temperature, annual precipitation, and humidity. 2. The dew point is the temperature at which water vapor starts to condense. 3. Relative humidity may decrease if water vapor condenses out of the air or if the air temperature decreases. 4. The type of cloud that forms at high altitudes is cirrocumulus. 5. The type of cloud that forms when strong air currents carry warm air upward is cumulus. 6. The type of fog that forms when cool air moves over a warm lake is steam fog. 7. Rain that passes through a layer of freezing air near the ground becomes sleet, which is different from hail and snow.
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1. If a radioactive sample has an initial count rate of 600 Bq. What is its count
rate after
i) l half-life?
ii) 2 half-lives
iii) 3 half-lives
iv) 4 half-lives?
Answer:
See explanation
Explanation:
Since the original count rate is 600 Bq,
i) after 1 half life, the count rate decreases to 1/2 of 600 Bq = 300 Bq
ii) after 2 half lives, the count rate decreases to 1/4 of 600 = 150 Bq
iii) after 3 half lives, the count rate decreases to 1/6 of 600 = 100 Bq
iv) after 4 half lives, the count rate decreases to 1/8 of 600 = 75 Bq
4.
How does the United Nations Development Program use its resources?
It provides health programs for mothers and children.
O It develops natural resources.
It works to eliminate poverty through development.
O It invests funds in industrialized nations.
Economic
Answer:
It develops natural resources.
It works to eliminate poverty through development.
If a car has a momentum of 2.04 x 104 kgm/s and a velocity of 18 m/s, what is its mass?
Answer:
Mass = 1133.33 kg (Approx.)
Explanation:
Given:
Momentum = 2.04 x 10⁴ kg[m/s]
Velocity = 18 m/s
Find:
Mass
Computation:
Mass = Momentum / Velocity
Mass = [2.04 x 10⁴] / 18
Mass = 1133.33 kg (Approx.)
if a current of 5A flows for 2minutes, find the quantity of electricity transfered
Define Refraction and give some knowlegde about it
To understand the nature of electric fields and how to draw field lines. Electric field lines are a tool used to visualize electric fields.
a. True
b. False
Answer:
a. True
Explanation:
Electric field is a region of space where the effect of electric field lines or lines of forces are felt.
The electric field lines creates electric field and these field lines help to visualize the electric field.
Therefore, electric field lines are tool used to visualize electric fields.
a. True
From rest, we step on the gas of our Ferrari, providing a force F for 4secs, speed up to a final speed v. If the applied force were only 1/2 F, how long would it have to be applied to reach the same final speed?
Answer:
The force must be applied during 8 seconds to reach trhe same final speed.
Explanation:
By Impulse Theorem, a change in the magnitude of linear momentum of a system with constant mass can be done by applying a force during a given time. That is:
[tex]m\cdot (v_{f}-v_{o}) = F \cdot \Delta t[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speed, measured in meters per second.
[tex]F[/tex] - Net external foce, measured in newtons.
[tex]\Delta t[/tex] - Time, measured in seconds.
We can eliminate mass and speeds by constructing the following relationship:
[tex]F_{1}\cdot \Delta t_{1} = F_{2}\cdot \Delta t_{2}[/tex] (2)
If we know that [tex]F_{1} = F[/tex], [tex]\Delta t_{1} = 4\,s[/tex] and [tex]F_{2} = \frac{F}{2}[/tex], then the time is:
[tex]4\cdot F = \frac{F\cdot \Delta t_{2}}{2}[/tex]
[tex]\Delta t_{2} = 8\,s[/tex]
The force must be applied during 8 seconds to reach trhe same final speed.
What type of bond is CO2?||
Answer:
Lol
Explanation:
