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andreyandreev [35.5K]
3 years ago
14

Suppose that a nascar race car is moving to the right with a constant velocity of +82m/s. what is the average acceleration of th

e car?
Physics
1 answer:
Veseljchak [2.6K]3 years ago
8 0
  <span>Iacceleration which would be roughly 16.17m/s^2</span><span>
</span>
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While traveling along a highway a driver slows from 24 m/sec to 15 m/sec in 12 seconds. What is the
Vladimir [108]
24-15=9 m/s slower in 12 seconds. So 9/12 m/s² slower. Therefore the acceleration is -0,75 m/s²
4 0
3 years ago
The speed of the center of the earth as it orbits the sun is 107257 kmph and the absolute angular velocity of the earth about it
Sindrei [870]

Answer with Explanation:

We are given that

Speed ,v_0=107257 kmph

Angular velocity,\omega=7.292\times 10^{-5} rad/s

Radius of earth,r=6371 km=6371000 m

1 km=1000 m

Linear velocity,v=r\omega=6371000\times 7.292\times 10^{-5}=464.57 m/s

Linear velocity,v=464.57\times \frac{18}{5}=1672.46 km/h

Velocity at point A,v_A=-vi+v_0 j=-1672.46 i+107257j kmph

Velocity at point B,v_B=v_0j-vj=107257j-1672.46j=105584.54j kmph

Velocity at point C,v_C=v_0j+vi=1672.46 i+107257j kmph

Velocity at point  D,v_D=v_0j+vj=107257j+1672.46j=108929.46jkmph

3 0
3 years ago
To minimize signal distortion, at each end of the J-1939 CAN-bus there is a(n)_____________resistor.
EastWind [94]
5-ohm
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120-ohm
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8 0
2 years ago
In which material does the light wave have the larger wavelength?
pochemuha

Answer: in the smaller/ thinner pieces.

8 0
2 years ago
A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon
Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force F_p by the worker must be equal to the friction force F_f on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N

b. The work is done on the crate by this force is the product of its force F_p and the distance traveled s = 4.35

W_p = F_ps = 79.1*4.35 = 344 J

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

W_f = F_fs = -79.1*4.35 = -344 J

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

W_p + W_f = 344 - 344 = 0 J

8 0
3 years ago
Read 2 more answers
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