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Black_prince [1.1K]
2 years ago
6

A vertical spring launcher is attached to the top of a block and a ball is placed in the launcher, as shown in the figure. While

the block slides at constant speed to the right across a horizontal surface with negligible friction between the block and the surface, the ball is launched upward. When the ball reaches its maximum height, what will be the position of the ball relative to the spring launcher?.
Physics
1 answer:
kvasek [131]2 years ago
7 0

For a vertical spring launcher is attached to the top of a block and a ball is placed in the launcher, the position of the ball will be  behind the box

<h3>What will be the position of the ball relative to the spring launcher?</h3>

Generally, the equation for the  conservation of momentum principle  is mathematically given as

(M+m) V1 = M*V2

Therefore, with the ball moving forward we have that; the ball at top it wii be behind the box,

Read more about Motion

brainly.com/question/605631

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A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr
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The  spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
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In a spring-mass system, the maximum velocity of the object is given by
v_{max} = A \omega
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v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s
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