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Reptile [31]
2 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]2 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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soldi70 [24.7K]

The entire motion of an object, regardless of direction.

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6 0
2 years ago
Each of the two coils has 320 turns. The average radius of the coil is 6 cm. The distance from the center of one coil to the ele
Temka [501]

Answer:

Magnetic field, B=2.39\times 10^{-3}\ T

Explanation:

It is given that,

Number of turns, N = 320

Radius of the coil, r = 6 cm = 0.06 m

The distance from the center of one coil to the electron beam is 3 cm, x = 3 cm = 0.03 m

Current flowing through the coils, I = 0.5 A

We need to find the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils. The magnetic field midway between the coils is given by :

B=\dfrac{\mu_oINr^2}{(x^2+r^2)^{3/2}}

B=\dfrac{4\pi \times 10^{-7}\times 0.5\times 320\times (0.06)^2}{(0.03^2+0.06^2)^{3/2}}

B = 0.00239 T

or

B=2.39\times 10^{-3}\ T

So, the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils is 2.39\times 10^{-3}\ T. Hence, this is the required solution.

7 0
4 years ago
<img src="https://tex.z-dn.net/?f=%5Chuge%5Cbold%7B%5Cpurple%7B%5Cbold%7B%E2%9A%A1Gravitational%20Constant%3F%E2%9A%A1%7D%7D%7D%
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\huge\underline\mathtt\colorbox{cyan}{G=}

6.673 \times  {10}^{ - 11}

And unit is Nm^2/kg^2

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Explanation:

i know the anwser

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