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Dmitriy789 [7]
3 years ago
11

A car is moving with a velocity of45m/s. Is brought to rest in 5s.the distance travelled by car before it comes to rest is

Physics
1 answer:
NikAS [45]3 years ago
6 0

Answer:

The car travels the distance of 225m before coming to rest.

Explanation:

Here,

v = 45m/s

t = 5s

d = v × t

Therefore,

d = 45 × 5

= 225m

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A 70.0-kg skier is sliding at 4 m/s when they slide down a 2m high hill. At the bottom of the hill they run into a large 2800 N/
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Answer:

\Delta x=245\ mm

Explanation:

Given:

  • mass of skier, m=70\ kg
  • initial velocity of skier, u=4\ m.s^{-1}
  • height of the hill, h=2\ m
  • spring constant, k=2800\ N.m^{-1}

<u>final velocity of skier before coming in contact of spring:</u>

Using eq. of motion:

v^2=u^2+gh

v^2=4^2+9.8\times 2

v=5.9666\ m.s^{-1}

<u>Now the time taken by the skier to reach down:</u>

v=u+gt

5.9666=4+9.8\ t

t=0.2007\ s

<u>Now we calculate force using Newton's second law:</u>

F=\frac{dp}{dt}

F=\frac{m(v-u)}{t}

F=\frac{70\times(5.9666-4)}{0.2007}

F\approx686\ N

<u>∴Compression in spring before the skier momentarily comes to rest:</u>

\Delta x=\frac{F}{k}

\Delta x=\frac{686}{2800}

\Delta x=0.245\ m

\Delta x=245\ mm

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