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2 years ago

A 1250-kg car moves at 20.0 m/s. How much work must be done on the car to increase its speed to 30.0 m/s.

Physics

1 answer:

bulgar [2K]2 years ago

6 0

Answer:

312,497.5Joules

Explanation:

Work done = force × distance

W = FS

Get the force

F = ma

F = 1250×9.8

F = 12250N

Get the distances using the equation of motion

v² = u² +2gS

30² =20²+2(9.8)S

900 =400+19.6S

900-400 =19.6S

500 = 19.6S

S = 500/19.6

S = 25.51m

Get the work done

Work done = 12250×25.51

Workdone = 312,497.5Joules

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2 years ago

A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i

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Answer:

a) Acceleration of the car is given as

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b) Acceleration of the truck is given as

$a_{truck} = 10.15 m/s^2$

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

$m_1v_1 + m_2v_2 = (m_1 + m_2)v$

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b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

$a_{truck} = \frac{v_f - v_i}{t}$

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