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slava [35]
2 years ago
13

The speed of a moving bullet can be deter-

Physics
1 answer:
Bad White [126]2 years ago
7 0

Answer:

<em>v = 381 m/s</em>

Explanation:

<u>Linear Speed</u>

The linear speed of the bullet is calculated by the formula:

\displaystyle v=\frac{x}{t}

Where:

x = Distance traveled

t = Time needed to travel x

We are given the distance the bullet travels x=61 cm = 0.61 m. We need to determine the time the bullet took to make the holes between the two disks.

The formula for the angular speed of a rotating object is:

\displaystyle \omega=\frac{\theta}{t}

Where θ is the angular displacement and t is the time. Solving for t:

\displaystyle t=\frac{\theta}{\omega}

The angular displacement is θ=14°. Converting to radians:

\theta=14*\pi/180=0.2443\ rad

The angular speed is w=1436 rev/min. Converting to rad/s:

\omega = 1436*2\pi/60=150.3776\ rad/s

Thus the time is:

\displaystyle t=\frac{0.2443\ rad}{150.3776\ rad/s}

t = 0.0016 s

Thus the speed of the bullet is:

\displaystyle v=\frac{0.61}{0.0016}

v = 381 m/s

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Answer: 80J

Explanation:

According to the first principle of thermodynamics:  

<em>"Energy is not created, nor destroyed, but it is conserved."  </em>

Then this priciple (also called Law) relates the work and the transferred heat exchanged in a system through the internal energy U, which is neither created nor destroyed, it is only transformed. So, in this especific case of the compressed gas:

\Delta U=Q+W  (1)

Where:

\Delta U is the variation in the internal (thermal) energy of the system (the value we want to find)

Q=-100J is the heat transferred out of the gas (that is why it is negative)

W is the work is done on the gas (as the gas is compressed, the work done on the gas must be considered positive )

On the other hand, the work done on the gas is given by:

W=-P \Delta V  (2)

Where:

P=450kPa=450(10)^{3}Pa is the constant pressure of the gas

\Delta V=V_{f}-V_{i} is the variation in volume of the gas

In this case the initial volume is V_{i}=600{cm}^{3}=600(10)^{-6}m^{3} and the final volume is V_{f}=200{cm}^{3}=200(10)^{-6}m^{3}.

This means:

\Delta V=200(10)^{-6}m^{3}-600(10)^{-6}m^{3}=-400(10)^{-6}m^{3}  (3)

Substituting (3) in (2):

W=-450(10)^{3}Pa(-400(10)^{-6}m^{3})  (4)

W=180J  (5)

Substituting (5) in (1):

\Delta U=-100J+180J  (6)

Finally:

\Delta U=80J  This is the change in thermal energy in the compression process.

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