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3 years ago

How to solve these two questions ?

Physics

1 answer:

murzikaleks [220]3 years ago

8 0

1) See attached graph

To solve this part of the problem, we have to keep in mind the relationship between current and charge:

$i = \frac{\Delta Q}{\Delta t}$

where

i is the current

Q is the charge

t is the time

The equation then means that the current is the rate of change of charge over time.

Therefore, if we plot a graph of the charge vs time (as it is done here), the current at any time will be equal to the slope of the charge vs time graph.

Here we have:

- Between t = 0 and t = 2 s, the slope is $\frac{50-0}{2-0}=25 C/s$ , so the current is 25 A

- Between t = 2 s and t = 6 s, the slope is $\frac{-50-(50)}{6-2}=-25 C/s$ , so the current is -25 A

- Between t = 6 s and t = 8 s, the slope is $\frac{0-(-50)}{8-6}=25 C/s$ , so the current is 25 A

Plotting on a graph, we find the graph in attachment.

2) $15 \mu C$

The relationship we have written before

$i = \frac{\Delta Q}{\Delta t}$

Can be rewritten as

$\Delta Q = i \Delta t$

This is valid for a constant current: if the current is not constant, then the total charge is simply equal to the area under the current vs time graph.

Therefore, we need to find the area under the graph.

Here we have a trapezium, where the two bases are

A = 1 ms = 0.001 s

B = 2 ms = 0.002 s

And the height is

h = 10 mA = 0.010 A

So, the area is

$Area=\frac{(0.001+0.002)\cdot 0.010}{2}=1.5\cdot 10^{-5} C = 15 \mu C$

So, the charge is $15 \mu C$ .

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Answer:

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

Explanation:

By the Principle of Energy Conservation and the Work-Energy Theorem we know that flywheel slow down due to the action of non-conservative forces (i.e. friction), the energy losses are equal to the change in the rotational kinetic energy. That is:

$\Delta E = K_{1}-K_{2}$ (1)

Where:

$\Delta E$ - Energy losses, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final rotational kinetic energies, measured in joules.

By definition of rotational kinetic energy, we expand the equation above:

$\Delta E = \frac{1}{2}\cdot I\cdot (\omega_{1}^{2}-\omega_{2}^{2})$ (2)

Where:

$I$ - Moment of inertia of the flywheel, measured in kilograms per square meter.

$\omega_{1}$ , $\omega_{2}$ - Initial and final angular speed, measured in radians per second.

If we know that $K_{1} = 30\,J$ , $K_{2} = 15\,J$ and $I = 18\,kg\cdot m^{2}$ , then the initial angular speed is:

$K_{1} = \frac{1}{2}\cdot I \cdot \omega_{1}^{2}$ (3)

$\omega_{1}=\sqrt{\frac{2\cdot K_{1}}{I} }$

$\omega_{1} = \sqrt{\frac{2\cdot (30\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{1} \approx 1.825\,\frac{rad}{s}$

$\omega_{1}\approx 0.291\,\frac{rev}{s}$

$K_{2} = \frac{1}{2}\cdot I \cdot \omega_{2}^{2}$ (4)

$\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }$

$\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{2} \approx 1.291\,\frac{rad}{s}$

$\omega_{2} \approx 0.205\,\frac{rev}{s}$

Under the assumption that flywheel is decelerating uniformly, we get that the time taken for the flywheel to slowdown is:

$t = \frac{\omega_{2}-\omega_{1}}{\alpha}$ (5)

If we know that $\omega_{1}\approx 0.291\,\frac{rev}{s}$ , $\omega_{2} \approx 0.205\,\frac{rev}{s}$ and $\alpha = -0.200\,\frac{rev}{s^{2}}$ , then the time needed is:

$t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }$

$t = 0.43\,s$

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

6 0

3 years ago

How to solve these two questions ?​

How to solve these two questions ?