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Vika [28.1K]
3 years ago
12

A 5kg cart moving to the right with a velocity of 16 m/s collides with a concrete wall and

Physics
1 answer:
Kryger [21]3 years ago
8 0

Explanation:

mass, m = 5kg

initial velocity, u = 16m/s

final velocuty, v = -22m/s

change in momentum, ∆p = ?

∆p = m (v-u)

5(-22-16)

5(38)

∆p = 190kgm/s

check the calculations!

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The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the
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Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

s=ut+0.5at^2

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, a=-10.0 m/s^2

\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is s=44.0 m

The acceleration is same, a=-10.0 m/s^2

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

v^2-u^2=2as

0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.





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Define rotational speed?- short answer
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A ball with an initial velocity of zero is dropped from a table onto the ground. After 2 s, the
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