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kap26 [50]
3 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]3 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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If you had started with a larger mass, how would the half-life change?
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A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se
zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

\vec{E'} = 1800\ N/C     (towards)

Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

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3 years ago
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