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dlinn [17]
3 years ago
9

A spring that has a force constant of 1050 N/m is mounted vertically on the ground. A block of mass 1.95 kg is dropped from rest

from height of 1.75 m above the free end of the spring. By what distance does the spring compress?
Physics
1 answer:
aleksandrvk [35]3 years ago
8 0

Answer:

25.2 cm

Explanation:

K = 1050 N/m

m = 1.95 kg

h = 1.75 m

By the conservation of energy, the potential energy of the block is converted into the potential energy stored in the spring

m g h = 1/2 x k x y^2

Where, y be the distance by which the spring is compressed.

1.95 x 9.8 x 1.75 = 1/2 x 1050 x y^2

33.44 = 525 x y^2

y = 0.252 m

y = 25.2 cm

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A circular orbit would have an eccentricity of
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6 0
3 years ago
8. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott
nasty-shy [4]

<u>Answer:</u> The ball is travelling with a speed of 5.5 m/s after hitting the <u>bottle.</u>

<u>Explanation:</u>

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m_1,u_1\text{ and }v_1 are the mass, initial velocity and final velocity of ball.

m_2,u_2\text{ and }v_2 are the mass, initial velocity and final velocity of bottle.

We are given:

m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s

Putting values in above equation, we get:

(0.4\times 18)+(0.2\times 0)=(0.4\times v_1)+(0.2\times 25)\\\\v_1=5.5m/s

Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

5 0
3 years ago
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