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Aleksandr [31]
3 years ago
15

A spring with k = 500 N/m stores 704 J. How far is it extended from the equilibrium position

Physics
1 answer:
kaheart [24]3 years ago
4 0

Given that:

k = 500 n/m,

work (W) = 704 J

spring extension (x) = ?

         we know that,

                      Work = (1/2) k x²

                          704 = (1/2) × 500 × x²

                            x = 1.67 m

A spring stretched for 1.67 m distance.

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8 0
3 years ago
A 328-kg car moving at 19.1 m/s in the x direction hits from behind a second car moving at 13.0 m/s in the same direction. If th
Jet001 [13]

Answer:

14.04 m/s

Explanation:

To find the velocity of the first car after the collision, we can use the equation of conservation of momentum:

m1v1 + m2v2 = m1'v1' + m2'v2'

We have the following data:

m1 = m1' = 328,

m2 = m2' = 790,

v1 = 19.1,

v2 = 13,

v2' = 15.1.

Using this data, we can find v1' (final velocity of the first car):

328 * 19.1 + 790 * 13 = 328 * v1' + 790 * 15.1

16534.8 = 328 * v1' + 11929

328 * v1' = 4605.8

v1' = 14.04 m/s

3 0
3 years ago
A box rests on the back of a truck. The coefficient of friction between the box and the surface is 0.32. (3 marks) a. When the t
cricket20 [7]

Answer:

Part a)

here friction force will accelerate the box in forward direction

Part b)

a = 3.14 m/s/s

Explanation:

Part a)

When truck accelerates forward direction then the box placed on the truck will also move with the truck in same direction

Here if they both moves together with same acceleration then the force on the box is due to friction force between the box and the surface of the truck

So here friction force will accelerate the box in forward direction

Part b)

The maximum value of friction force on the box is known as limiting friction

it is given by the formula

F = \mu mg

so we have

F = ma = \mu mg

now the acceleration is given as

a = \mu g

a = (0.32)(9.8) = 3.14 m/s^2

5 0
3 years ago
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