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kap26 [50]
2 years ago
8

We have a toy gun with a spring constant of 50 N/m. The spring is compressed by 0.2 m. If you neglect friction and the mass of t

he spring, at what speed will a 2 g projectile be ejected from the gun
Physics
1 answer:
Arisa [49]2 years ago
4 0

Answer:

31.6\:\mathrm{m/s}

Explanation:

The elastic potential energy of a spring is given by Us=\frac{1}{2}kx^2, where k is the spring constant of the spring and x is displacement from point of equilibrium.

When released, this potential energy will be converted into kinetic energy. Kinetic energy is given by KE=\frac{1}{2}mv^2, where m is the mass of the object and v is the object's velocity.

Thus, we have:

Us=KE,\\\frac{1}{2}kx^2=\frac{1}{2}mv^2

Substituting given values, we get:

\frac{1}{2}\cdot 50\cdot 0.2^2=\frac{1}{2}\cdot 0.002\cdot v^2,\\v^2=\frac{50\cdot 0.2^2}{0.002},\\v^2=1000,\\v\approx \boxed{31.6\:\mathrm{m/s}}

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Answer: Current = 2 A

Explanation:

Given that an electrical power plant generates electricity with a

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The resistance R will be achieved by Ohms law formula which state that

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Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8
umka2103 [35]

Answer:

a) F=2.048\times 10^{-7}\ N

b) a=0.1138\ m.s^{-2}

Explanation:

Given:

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  • charge on the raindrops, q=+21\times 10^{-12}\ C
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A)

<u>From the Coulomb's Law the force between the charges is given as:</u>

F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}

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<em>Now force:</em>

F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}

F=2.048\times 10^{-7}\ N

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a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}

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