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Margaret [11]
3 years ago
14

A car achieves a velocity 50 meters per second after 5 seconds. What is the Car's acceleration?

Physics
1 answer:
Reptile [31]3 years ago
7 0

Answer: 10 m/s

Explanation: Velocity/Time

50/5= 10

:)

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Pressing two objects together with more force __________ friction
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Pressing two objects together with more force Increase friction
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The planar simple harmonic wave travels in the positive direction of x axis with wave velocity u=2m/s, and the vibration curve o
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The planar simple harmonic wave travels in the positive direction of x axis with wave velocity u=2m/s, and the vibration curve of the particle at the origin in cosinusoidal form is shown in the figure.

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7 0
2 years ago
Read 2 more answers
A railroad freight car, mass 15,000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples w
gayaneshka [121]

Answer:

The speed of the two cars after coupling is 0.46 m/s.

Explanation:

It is given that,

Mass of car 1, m₁ = 15,000 kg

Mass of car 2, m₂ = 50,000 kg

Speed of car 1, u₁ = 2 m/s

Initial speed of car 2, u₂ = 0

Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

V=\dfrac{15000\ kg\times 2\ m/s+0}{(15000\ kg+50000\ kg)}  

V = 0.46 m/s

So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.          

3 0
3 years ago
Your friend decides to generate electrical power by rotating a 100,000 turn coil of wire around an axis in the plane of the coil
MakcuM [25]

Answer:

a) I=35mA

b) P=1.73W

Explanation:

a) The max emf obtained in a rotating coil of N turns is given by:

emf_{max}=NBA\omega

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

A=\pi r^2 =\pi(0.23m)^2=0.16m^2

emf_{max}=(100000)(0.3*10^{-4}T)(0.166m^2)(140\frac{rev}{s})=69.72V

Finally, the peak current is given by:

I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA

b)

we have that

I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A

P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W

hope this helps!!

6 0
3 years ago
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