Answer:
A) v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]
B)√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
C) µ_s = tan θ
D) µ_s = 0.4663
Explanation:
A) The forces acting on the car will be;
Force due to friction; F_f
Force due to Gravity; F_g
Normal Force; F_n
Now, let us take the vertical direction to be j^ and the direction approaching the centre to be i^ downwards and parallel to the road surface by k^.
Also, we will assume that initially, F_n is in the negative k^ direction and that it will have a maximum possible value of; F_f = µ_s × F_n
Thus, sum of forces about the vertical j^ direction gives;
ΣF_j^ = F_n•cos θ − mg + F_f•sin θ = 0
Since F_f = µ_s × F_n ;
F_n•cos θ − mg + (µ_s × F_n × sin θ) =0
F_n = mg/[cos θ + (µ_s•sin θ)]
Also, sum of forces about the centre i^ direction gives;
ΣF_i^ = F_n(sin θ + (µ_s•cos θ)) = mv²/r
Plugging in formula for F_n gives;
ΣF_i^ = [mg/[cos θ + (µ_s•sin θ)]] × (sin θ + (µ_s•cos θ)) = mv²/r
Making v the subject gives;
v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]
B) What we got in a above is the minimum speed the car can have while going round the turn.
The maximum speed will be gotten by making the frictional force(F_f) to point in the positive k^ direction. This means that F_f will be negative.
Now, if we change the sign in front of F_f in the equation in part a that led to the minimum velocity, we will have the maximum as;
v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
Thus the range is;
√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]
C) For the minimum speed to be 0, it implies that F_f will be in the negative k^ direction. Thus, Sum of the forces in the k^ direction gives;
ΣF_k^ = mg(sin θ - µ_s•cos θ) = 0
Thus;
mg(sin θ - µ_s•cos θ) = 0
Making µ_s the subject gives;
µ_s = sin θ/cos θ
µ_s = tan θ
D) If θ = 25.0°;
Thus;
µ_s = tan 25
µ_s = 0.4663
Two velocity vectors, one is twice of the other, and separated by 90 Degree angle. If their resultant is calculated 90 m/s, what the magnitude of bigger vector?
Answer:
Vy = 80.5 [m/s]
Explanation:
In order to solve this problem we must use the Pythagorean theorem.
V = 90 [m/s]
The components are Vx and Vy:
Therefore:
[tex]v=\sqrt{v_{x}^{2} + v_{y}^{2} }[/tex]
where:
Vy = 2*Vx ; because one is twice of the other.
[tex]90 = \sqrt{v_{x}^{2} +(2*v_{x})^{2} }\\ 90 =\sqrt{v_{x}^{2}+4*v_{x}^{2}} \\90 =\sqrt{5v_{x}^{2}} \\90=2.23*v_{x} \\v_{x}=40.25[m/s][/tex]
and the bigger vector is:
Vy = 40.25*2
Vy = 80.5 [m/s]
How long does it take a cheetah that runs with a velocity of 34m/s to run 750m?
Answer:
It would take 5 seconds
Explanation:
I can't find the 'delta' sign nor the vector sign so just pretend that displacement, time and velocity has them.
V = d / t
34 = 170 / t
34 x t = 170
34t = 170
t = 170 / 34
t = 5
A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 126 m/s from the top of a cliff 132 m above level ground, where the ground is taken to be y = 0. (a) What is the initial total mechanical energy of the projectile? (Give your answer to at least three significant figures.) J (b) Suppose the projectile is traveling 89.3 m/s at its maximum height of y = 297 m. How much work has been done on the projectile by air friction? J (c) What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up?
Answer:
a) The initial total mechanical energy of the projectile is 498556.296 joules.
b) The work done on the projectile by air friction is 125960.4 joules.
c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.
Explanation:
a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ([tex]E[/tex]) of the project is equal to the sum of gravitational potential energy ([tex]U_{g}[/tex]) and translational kinetic energy ([tex]K[/tex]), all measured in joules:
[tex]E = U_{g} + K[/tex] (Eq. 1)
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the projectile, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y[/tex] - Initial height of the projectile above ground, measured in meters.
[tex]v[/tex] - Initial speed of the projectile, measured in meters per second.
If we know that [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y = 132\,m[/tex] and [tex]v = 126\,\frac{m}{s}[/tex], the initial mechanical energy of the earth-projectile system is:
[tex]E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}[/tex]
[tex]E = 498556.296\,J[/tex]
The initial total mechanical energy of the projectile is 498556.296 joules.
b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:
[tex]W_{loss} = E_{o}-E_{1}[/tex] (Eq. 2)
Where:
[tex]E_{o}[/tex] - Initial total mechanical energy, measured in joules.
[tex]E_{1}[/tex] - FInal total mechanical energy, measured in joules.
[tex]W_{loss}[/tex] - Work losses due to air friction, measured in joules.
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]W_{loss} = E_{o}-K_{1}-U_{g,1}[/tex]
[tex]W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}[/tex] (Eq. 2b)
Where:
[tex]m[/tex] - Mass of the projectile, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex] - Maximum height of the projectile above ground, measured in meters.
[tex]v_{1}[/tex] - Current speed of the projectile, measured in meters per second.
If we know that [tex]E_{o} = 498556.296\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 297\,m[/tex] and [tex]v_{1} = 89.3\,\frac{m}{s}[/tex], the work losses due to air friction are:
[tex]W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)[/tex]
[tex]W_{loss} = 125960.4\,J[/tex]
The work done on the projectile by air friction is 125960.4 joules.
c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:
[tex]E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}[/tex] (Eq. 3)
[tex]K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}[/tex]
Where:
[tex]E_{1}[/tex] - Total mechanical energy of the projectile at maximum height, measured in joules.
[tex]U_{g,2}[/tex] - Potential gravitational energy of the projectile, measured in joules.
[tex]K_{2}[/tex] - Kinetic energy of the projectile, measured in joules.
[tex]W_{loss}[/tex] - Work losses due to air friction during the upward movement, measured in joules.
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}[/tex] (Eq. 3b)
[tex]m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}[/tex]
[tex]v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}[/tex]
[tex]v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }[/tex]
If we know that [tex]E_{1} = 372595.896\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{2} =0\,m[/tex] and [tex]W_{loss} = 125960.4\,J[/tex], the final speed of the projectile is:
[tex]v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }[/tex]
[tex]v_{2} \approx 82.475\,\frac{m}{s}[/tex]
The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.
A sound wave in a steel rail has a frequency of 620 hz and a wavelength of 10.5m. What is the speed of sound in steel?
The average speed and kinetic energy of the particles in a gas are proportional to the measured what of the gas
Answer: Temperature
Explanation:
The average kinetic energy of the particles in a gas is proportional to the temperature of the gas. Because the mass of these particles is constant, the particles must move faster as the gas becomes warmer.
What is kinetic energy?
"The kinetic energy of an object is the energy that it possesses due to its motion."
What is average kinetic energy?
"The product of the half of the mass of each gas molecule and the square of RMS speed."
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Help pls it’s urgent
write .00000023 in scientific notation
Answer: 2.3 x 10^-7
Explanation:
When you move the decimal to the right, you decrease the exponent. So lets count how many spots you have to move the decimal spot to get an answer between 1 and 10. I counted seven spots.
So:
[tex]2.3 *10^-^7[/tex]
2.3 * 10 °−7
The degree indicates that the number is that small thing that floats above
How can models help us understand energy?
Black and splinter cleavage barely scratches glass
Answer:
oh I know that sounds good to me
If you want to make a strong battery, should you pair two metals with high electron affinities, low electron affinities, or a mix? Explain your answer.
A worker pushes on a crate, and it experiences a net force of
300 N. If the crate moves with an acceleration of 0.750 m/s2
,
what is its mass?
We are given:
F = 300N
acceleration (a) = 0.750 m/s²
Solving for the mass of the object:
We know from the second equation of motion:
F = ma
replacing the variables
300 = m * 0.75
m = 300 * 100/ 75
m = 300 * 4/3
m = 100 * 4
m = 400 kg
10. A boy runs 5 miles East then turns around and runs 7.5 miles West. What is
his displacement?
Answer:
2.5
Explanation:
When a space shuttle was launched, the astronauts onboard experienced an acceleration of 32.0 m/s2 . If one of the astronauts had a mass of 40.0 kg, what net force in newtons did the astronaut experience?
Answer:
F = 1280 N
Explanation:
Given that,
Acceleration experienced by a space shuttle, a = 32 m/s²
Mass of the astronauts, m = 40 kg
We need to find the force experienced by the astronaut.
We know that the net force is equal to the product of acceleration and its mass. So,
F = ma
F = 40 kg × 32 m/s²
So,
F = 1280 N
So, 1280 N of force is experienced by the Astronaut.
How much distance did this object travel in meters between 0 and 10 seconds
Answer:
5m
Explanation:
Answer:
5
Explanation:
because that's the average of how far it when the most
according to newton’s first law a moving object acted on by a net force of zero....
Answer:
Newton's first law says that if the net force on an object is zero ( Σ F = 0 \Sigma F=0 ΣF=0\Sigma, F, equals, 0), then that object will have zero acceleration. That doesn't necessarily mean the object is at rest, but it means that the velocity is constant.
Explanation:
According to Newton’s first law of motion, a moving object acted on by a net force of zero will continue in motion.
Newton's first law of motion states that an object in state of rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force.
This first is also known as the law of inertia because it is the reluctance of an object in motion to stop moving or reluctance of an object at rest to start moving which depends on the mass of the object.
Thus, we can conclude that according to Newton’s first law of motion, a moving object acted on by a net force of zero will continue in motion.
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1. What happens to the current in a series circuit as it moves through each component? a. The current stays the same throughout the circuit.
b. The current will increase or decrease depending on the resistance.
c. The current decreases with each component it goes through.
d. The current increases with each component it goes through.
2. A 10-volt power supply is placed in series with two 5-ohm resistors. What is the current in the circuit after it passes through each of the two resistors?(1 point)
a. The current will stay the same at 1 amp after passing through both resistors.
b. The current will drop to 2 amps after the first resistor and then to 1 amp after the second resistor.
c. The current will stay the same at 2 amps after passing through both resistors.
d. The current will drop to 1 amp after the first resistor and then to 0 amps after the second resistor.
3. What is the voltage that passes through R1 and R2?
a. R1: 12 V, R2: 24 V
b. R1: 8 V, R2: 4 V
c. R1: 12 V, R2: 12 V
d. R1: 6 V, R2: 6 V
4. Which of the following correctly describes the magnitude of currents I1 and I2 ?
a. I1 is equal to I2
b. I1 and I2 approach zero
c. I1 is greater than I2
d. I1 is less than I2
5. If the energy of an electric charge flowing in a circuit is conserved, which of the following obeys the Kirchhoff junction rule?
a. The sum of the current flowing in is greater than the sum of current flowing out.
b. The sum of the current flowing in is less than the sum of the current flowing out.
c. The sum of the current flowing in is equal to the sum of current flowing out.
d. The sum of the current flowing in is zero and the sum of the current flowing out is greater than zero.
Answer: sorry here’s the answers, I didn’t feel like typing it all
Explanation:
The correct answer to the 5 questions are;
1) Option A; The current stays the same throughout the circuit.
2) Option A; The current will stay the same at 1 amp after passing through both resistors.
3) Option C; R1: 12 V, R2: 12 V
4) Option A; I1 is equal to I2
5) Option C; The sum of the current flowing in is equal to the sum of current flowing out.
1) In an electrical circuit, usually as current moves through each component, it stays the same. Thus, option A is correct.
2) Formula for current is;
I = V/R
We are told that there are two resistors in series each having a resistance of 5Ω. Thus; Total resistance = 5 + 5 = 10 Ω.
Thus; Current = 10/10 = 1 A.
The current will stay same at 1 A after passing through both resistors.
3) From the circuit we are given, we see that the Voltage is 12 V. Now, the same voltage would be transmitted through both resistor R1 and R2.
Option C is correct
4) The current splits upon passing resistor 1 and as such it means the current I2 going through the second resistor would be the same. Thus; I1 = I2.
5) Kirchoff's junction rule states that all the incoming currents to a particular junction must be equal to sum of all currents going out of that same junction. Thus, option C is correct.
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Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventually be converted into sweat and evaporate. If you drink a 20.0-ounce bottle of water that had been in the refrigerator at 3.8 °C, how much heat is needed to convert all of that water into sweat , knowing 1ml contains 0.03 ounces? (Note: Your body temperature is 36.6 °C. For the purpose of solving this problem, assume that the thermal properties of sweat are the same as for water.)
Answer:
The amount of heat required is [tex]H_t = 1.37 *10^{6} \ J [/tex]
Explanation:
From the question we are told that
The mass of water is [tex]m_w = 20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g[/tex]
The temperature of the water before drinking is [tex]T_w = 3.8 ^oC[/tex]
The temperature of the body is [tex]T_b = 36.6^oC[/tex]
Generally the amount of heat required to move the water from its former temperature to the body temperature is
[tex]H= m_w * c_w * \Delta T[/tex]
Here [tex]c_w [/tex] is the specific heat of water with value [tex]c_w = 4.18 J/g^oC [/tex]
So
[tex]H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)[/tex]
=> [tex]H= 7.8 *10^{4} \ J [/tex]
Generally the no of mole of sweat present mass of water is
[tex]n = \frac{m_w}{Z_s}[/tex]
Here [tex]Z_w[/tex] is the molar mass of sweat with value
[tex]Z_w = 18.015 g/mol[/tex]
=> [tex]n = \frac{5.7 *10^2}{18.015}[/tex]
=> [tex]n = 31.6 \ moles [/tex]
Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as
[tex]H_v = n * L_v[/tex]
Here [tex]L_v[/tex] is the latent heat of vaporization with value [tex]L_v = 7 *10^{3} J/mol[/tex]
=> [tex]H_v = 31.6 * 7 *10^{3} [/tex]
=> [tex]H_v = 1.29 *10^{6} \ J [/tex]
Generally the overall amount of heat energy required is
[tex]H_t = H + H_v[/tex]
=> [tex]H_t = 7.8 *10^{4} + 1.29 *10^{6}[/tex]
=> [tex]H_t = 1.37 *10^{6} \ J [/tex]
David drove the first 6 hours of his journey at 65km/hr and the last 3 hours of his journey at 80km/hour. How far is the whole journey in km?
630 kilometers for the whole journey
Which parts of the warm-up did you find most difficult? Why?
Answer:
Can't really answer that for now. More context please?
Explanation:
I will answer it in a comment when you give some context.
A 30-cm-diameter, 4-m-high cylindrical column of a house made of concrete ( k = 0.79 W/m⋅K, α = 5.94 × 10 −7 m2/s, rho = 1600kg/ m 3 , and c p = 0.84kJ/kg⋅K ) cooled to 14° C during a cold night is heated again during the day by being exposed to ambient air at an average temperature of 28° C with an aver-age heat transfer coefficient of 14 W/ m 2 ⋅K. Using the analyti-cal one-term approximation method, determine (a) how long it will take for the column surface temperature to rise to 27° C, (b) the amount of heat transfer until the center temperature reaches to 28° C, and (c) the amount of heat transfer until the surface temperature reaches 27° C.
Answer:
a) Time it will taken for the column surface temperature to rise to 27°C is
17.1 hours
b) Amount of heat transfer is 5320 kJ
c) Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ
Explanation:
Given that;
Diameter D = 30 cm
Height H = 4m
heat transfer coeff h = 14 W/m².°C
thermal conductivity k = 0.79 W/m.°C
thermal diffusivity α = 5.94 × 10⁻⁷ m²/s
Density p = 1600 kh/m³
specific heat Cp = 0.84 Kj/kg.°C
a)
the Biot number is
Bi = hr₀ / k
we substitute
Bi = (14 W/m².°C × 0.15m) / 0.79 W/m.°C
Bi = 2.658
From the coefficient for one term approximate of transient one dimensional heat conduction The constants λ₁ and A₁ corresponding to this Biot number are,
λ₁ = 1.7240
A₁ = 1.3915
Once the constant J₀ = 0.3841 is determined from corresponding to the constant λ₁
the Fourier number is determined to be
[ T(r₀, t) -T∞ ] / [ Ti - T∞] = A₁e^(-λ₁²t') J₀ (λ₁r₀ / r₀)
(27 - 28) / (14 - 28) = (1.3915)e^-(17240)²t (0.3841)
t' = 0.6771
Which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the time it will take for the column surface temperature to rise to 27°C becomes
t = t'r₀² / ₐ
= (0.6771 × 0.15 m)² / (5.94 x 10⁻⁷ m²/s)
= 23,650 s
= 7.1 hours
Time it will taken for the column surface temperature to rise to 27°C is
17.1 hours
b)
The heat transfer to the column will stop when the center temperature of column reaches to the ambient temperature, which is 28°C.
Maximum heat transfer between the ambient air and the column is
m = pV
= pπr₀²L
= (1600 kg/m³ × π × (0.15 m)² × (4 m)
= 452.389 kg
Qin = mCp [T∞ - Ti ]
= (452.389 kg) (0.84 kJ/kg.°C) (28 - 14)°C
= 5320 kJ
Amount of heat transfer is 5320 kJ
(c)
the amount of heat transfer until the surface temperature reaches to 27°C is
(T(0,t) - T∞) / Ti - T∞ = A₁e^(-λ₁²t')
= (1.3915)e^-(1.7240)² (0.6771)
= 0.1860
Once the constant J₁ = 0.5787 is determined from Table corresponding to the constant λ₁, the amount of heat transfer becomes
(Q/Qmax)cyl = 1 - 2((T₀ - T∞) / ( Ti - T∞)) ((J₁(λ₁)) / λ₁)
= 1 - 2 × 0.1860 × (0.5787 / 1.7240)
= 0.875
Q = 0.875Qmax
Q = 0.875(5320 kJ)
Q = 4660 kJ
Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ
You have purchased an inexpensive USB oscilloscope (which measures and displays voltage waveforms). You wish to determine if the oscilloscope has an error bias; in other words, you wish to determine if the errors made by the oscilloscope have a population mean that is not equal to zero. So you use a very accurate voltmeter to find the measurement errors for 13 different measurements made by your USB oscilloscope. A data file containing these measurements is HTMean1.csvPreview the document . Do a statistical analysis on this data to determine if the oscilloscope has an error bias.
Complete Question
The complete question is shown on the first uploaded image
Answer:
1
A
2
A
Explanation:
From the question the data given for Error (mV) is -15
-15.17
8.67
-13.74
-20.69
-6.96
-1.36
-2.96
-9.26
3.11
-14.12
6.39
-14.77
Generally
The null hypothesis is [tex]H_o : \mu = 0[/tex]
The alternative hypothesis is [tex]H_a : \mu \ne 0[/tex]
The sample size is n = 13
Here [tex]\mu[/tex] represents the true error bias (i.e population error bias)
Generally the sample error bias is mathematically represented as
[tex]\= x = \frac{ \sum x_i}{n}[/tex]
=> [tex]\= x = \frac{ -15.17 + 8.67 + (-13.74) + \cdots + (-14.77) }{13}[/tex]
[tex]\= x = -7.37[/tex]
Generally the standard deviation is mathematically represented as
[tex]\sigma = \sqrt{\frac{\sum (x_i - \= x )^2}{n} }[/tex]
=> [tex]\sigma = \sqrt{\frac{ (-15.17-( -7.37) )^2 + (8.67 -( -7.37) )^2 + \cdots + (-14.77 -( -7.37) )^2 }{13} }[/tex]
=> [tex]\sigma = \sqrt{ 119.385}[/tex]
=> [tex]\sigma = 10.926[/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{\= x - \mu }{\frac{\sigma }{\sqrt{n} } }[/tex]
=> [tex]t = \frac{ -7.37 - 0 }{\frac{10.926}{\sqrt{13} } }[/tex]
=> [tex]t = -2.838[/tex]
Generally the p-value is mathematically represented as
[tex]p-value = 2 P(t < -2.432)[/tex]
From the z-table [tex]P(t < -2.432) = 0.0075 [/tex]
So [tex]p-value = 2* 0.0075 [/tex]
=> [tex]p-value = 0.015 [/tex]
So given that p-value is less than the [tex] \alpha = 0.05[/tex] then we reject the null hypothesis and conclude that the oscilloscope has an error bias
A car goes 500m in 5 sec. It was moving 10 m/sec to begin with, what is
its final velocity?*
Answer:
v = 190 m/s
Explanation:
Given that,
Initial velocity of a car, u = 10 m/s
Distance, d = 500 m
Time, t = 5 s
We need to find the final velocity of the car. Firstly we can find the acceleration of the car using second equation of motion as follows :
[tex]d=ut+\dfrac{1}{2}at^2\\\\500=10\times 5+\dfrac{a}{2}\times 5^2\\\\500=50+\dfrac{25a}{2}\\\\450=\dfrac{25a}{2}\\\\a=\dfrac{450\times 2}{25}\\\\a=36\ m/s^2[/tex]
let v is the final velocity. using First equation of motion to find a as follows :
v=u+at
v=10+36(5)
v=190 m/s
So, the final velocity of the car is 190 m/s.
____ can be calculated if you know the distance that an object travels in one unit of time.
A.motion
B.meter
C.Rate
D.Speed
E.velocity
F.slope
G.refrence point
PLS HELP NOW !!!
Answer:
D.Speed
Explanation:
The speed of an object is the distance the object travels in one unit of time.
Speed can be calculated if you know the distance that an object travels in one unit of time, therefore the correct answer is option D.
What is speed?The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object.
The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.
Thus, Speed can be calculated if you know the distance that an object travels in one unit of time, therefore the correct answer is option D.
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This chart shows Dan's budget:
Did Dan stay on budget? Why or why not?
Amount budgeted
tem
nome
Food
Rent
Debonary spending
Income
$100
S500
$100
5750
Amount spent
55
S90
3500
5140
Yes, Dan spent as much as he earned.
No, Dan should move to a new apartment
O Yes, Dan uses his savings to cover extra expense
No, Dan should reduce his discretionary spending,
Answer
No dan should reduce his discretionary apendings.
Explanation:
What are some examples of magmatism?
which of the following statements describes how a balance obtains a measurement
A. A ruler is placed next to an object, and its length is noted.
B. An object is placed on one side and weights are placed on the other side until the two dishes are even.
C. The sides of an object are measured and multiplied to get the final measurement.
D. An object is placed in a cylinder, and the water displaced gives the measurement.
Which of the following statements describes how a balance obtains a measurement
Explanation:
A balance obtains a measurement by the sides of an object which are measured and multiplied to get the final measurement. Thus, the correct option is C.
How a balance obtains measurement?A balance is an instrument used for comparing the weights of two objects. It is usually for the scientific purposes, to determine the difference in mass or weight of an object.
Weighing of objects directly requires that the balance be carefully zeroed that is it reads zero with nothing on the balance pan in order to obtain accurate results of the measurement.
The sides of an object are measured and multiplied with a factor to get the final measurement of the object. Through this, a balance obtains the measurement.
Therefore, the correct option is C.
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Write the equation for newtons third law
Answer:
Explanation:
Newtons third law says an applied force will produce an equal but opposite force.
[tex]F_A_B =-F_B_A[/tex]
Explain period motion of oscillatory body.
Answer:
motion repeating itself is referred to as periodic or oscillatory motion. An object in such motion oscillates about an equilibrium position due to a restoring force or torque. ... This motion is important to study many phenomena including electromagnetic waves, alternating current circuits, and molecules.
Explanation:
Answer:
Moves to and from about it's mean position in a fixed time interval.
increased force will increase acceleration true or false.
A hockey player whacks a 162-g puck with her stick, applying a 171-N force that accelerates it to 42.3 m/s. A. If the puck was initially at rest, for how much time did the acceleration last? B. The puck then hits the curved corner boards, which exert a 151-N force on the puck to keep it in its circular path. What’s the radius of the curve?
Given parameters:
Mass of puck = 162g = 0.162kg (1000g = 1kg)
Force exerted on puck = 171N
Final velocity = 42.3m/s
Unknown
A. time of the acceleration
B. radius of the curve?
Solution:
A. time of the acceleration
the initial velocity of the puck = 0m/s
We know that;
Force = mass x acceleration
Acceleration = [tex]\frac{Final velocity - Initial velocity}{time taken}[/tex]
Acceleration = [tex]\frac{42.3 - 0}{t}[/tex]
So force = mass x [tex]\frac{42.3 }{t}[/tex]
Input the parameters and solve for time;
171 = 0.162 x [tex]\frac{42.3 }{t}[/tex]
171 = [tex]\frac{6.85}{t}[/tex]
t = [tex]\frac{6.85}{171}[/tex] = 0.04s
The time of acceleration is 0.04s
B. radius of the curve;
to solve this, we apply the centripetal force formula;
F = [tex]\frac{mv^{2} }{r}[/tex]
where;
F is the centripetal force
m is the mass
v is the velocity
r is the radius
Since the force exerted on the puck is 151;
input the parameters and solve for r²;
151 = [tex]\frac{0.162 x 42.3^{2} }{r}[/tex]
151r = 0.162 x 42.3²
r = 1.92m
The radius of the circular curve is 1.92m
