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Charra [1.4K]
3 years ago
15

What is the acceleration of a car that maintains a constant velocity of 100 km/h for 10 sec?

Physics
1 answer:
MariettaO [177]3 years ago
8 0

It doesn't matter what the velocity is, or for how long.
They could both be anything.

"Constant velocity" means <em>zero</em> acceleration.


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A box with mass 25.14 kg is sliding at rest from the top of the slope with height 13.30 m
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Answer:

Explanation:

The first thing we are asked to find is the Force experienced by the box. That is found in the formula:

F - f = ma where F is the force exerted by the box, f is the friction opposing the box, m is the mass, and a is the acceleration (NOT the same as the pull of gravity). But F can be rewritten in terms of the angle of inclination also:

wsin\theta-f=ma where w is the weight of the box. We will use this version of the formula because it will help us answer the second question, which is to solve for a. Filling in:

First we need the weight of the box. Having the mass, we find the weight:

w = mg so

w = 25.14(10) so

w = 251.4 N (I am not paying any attention at all to the sig fig's here, since I noticed no one on this site does!) Now we have the weight. Filling that in:

251.4sin(30) - f = ma Before we go on to fill in for f, let's answer the first question. F = 251.4sin(30) so

F = 125.7   And in order to answer what a is equal to, we find f:

f = μF_n where Fn is the weight of the object.

f = .25(251.4) so

f = 62.85. Filling everything in now altogether to solve for a, the only missing value:

125.7 - 62.85 = 25.14a and

62.85 = 25.14a so

a = 2.5 m/s/s

Now we have to move on to another set of equations to answer the last part. The last part involves the y-dimension. In this dimension, what we know is that

a = -10 m/s/s

v₀ = 0 (it starts from rest)

Δx = -13.30 m (negative because the box falls this fr below the point fro which it started). Putting all that together in the equation for displacement:

Δx = v₀t + \frac{1}{2}at^2 and we are solving for time:

-13.30=0t+\frac{1}{2}(-10)t^2 and

t=\sqrt{\frac{2(-13.30)}{-10} } so

t = 1.6 seconds to reach the bottom of the slope from 13.30 m high.

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Answer / Explanation:

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<span>b. The passengers aboard on train A will see that train B is moving at 41m/sec due west</span>

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