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Reptile [31]
3 years ago
10

A spring with a spring constant value of 125 N/m is compressed 12.2 cm by pushing on it with a 215 g block. When the block is re

leased, what velocity will the block have when it leaves the spring (we're ignoring friction here)? a​
Physics
1 answer:
allsm [11]3 years ago
3 0

Answer:

v = 2.94 m/s

Explanation:

When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.

Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.

Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means

(1/2)kx^2 = (1/2)mv^2

kx^2 = mv^2

v^2 = (kx^2)/m

v = sqrt((kx^2)/m)

v = x * sqrt(k/m)

v = 0.122 * sqrt(125/0.215)        <--- units converted to m and kg

v = 2.94 m/s

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expeople1 [14]

Answer:

a) 1.68 N b) 0 c) 14.5º

Explanation:

a)

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  • where q= magnitude of the charge of the particle, v=velocity of the particle, B= magnitude of the magnetic field, and θ= angle between v and B.
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       F = q*v*B*sin\theta (1) = 6.00e-6C*1.40e5m/s*2T*sin\theta=\\ \\ F= 1.68N*sin\theta

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b)

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c)

  • If the force is 0.25 of the maximum possible, (which is when sin θ =1), this means that sin θ = 0.25, as it can be seen below:

       F_{\theta} = q*v*B*sin \theta = 0.25*Fmax\\ F_{\theta} = q*v*B*sin\theta= 0.25 (q*v*B*sin 90) \\ sin \theta = 0.25

       \theta = sin^{-1} (0.25) =14.5 deg

  • The angle between the velocity v and the magnetic field B is 14.5º.
3 0
4 years ago
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Answer:

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Explanation:

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Explanation:

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While the current in the wire produces a magnetic field and exerts a force on the needle. The insulation on the wire becomes energized and exerts a force on the needle. Hence, the compass needle moves and changes its position.

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