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OLEGan [10]
3 years ago
5

Sylvia and Jadon now want to work a problem. Imagine a puck of mass 0.5 kg moving as in the simulation. Suppose that the tension

in the string is 1.0 N, and that the radius of its circular path is 0.7 m. What will Jadon and Sylvia find for the tangential speed of the puck? m/s
Physics
1 answer:
Pavlova-9 [17]3 years ago
5 0

Answer:

1.2 m/s

Explanation:

The puck in this problem is moving with uniform circular motion, so the net force acting on it (the tension in the string) must be equal to the centripetal force.

So we can write:

T=m\frac{v^2}{r}

where:

T is the tension in the string

m is the mass of the puck

v is its tangential speed

r is the radius of the circular path

For the puck in this problem, we have:

m = 0.5 kg

T = 1.0 N

r = 0.7 m

Substituting and solving for v, we find the tangential speed:

v=\sqrt{\frac{Tr}{m}}=\sqrt{\frac{(1.0)(0.7)}{0.5}}=1.2 m/s

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Because plastic and rubber are insulators.
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How are a wave's energy and the wave's amplitude related?
dem82 [27]
Answer:
Energy is proportional to the square of the amplitude

Explanation:
The energy of a certain wave is defined using its magnitude.
The two quantities are related directly. This means that as the amplitude of the wave increases, its energy increases and vice versa.

Energy is directly proportional to the square of the magnitude of the wave. This means that:
If we have new amplitude = 2 * old amplitude
We will have new energy = (2)² * old energy = 4 * old energy

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7 0
3 years ago
A rotating object has an angular acceleration of α = 0 rad/s2. Which one or more of the following three statements is consistent
Murrr4er [49]

Answer:

A,B and C

Explanation:

Statement A  

At all times, angular velocity is \omega = 0\,{\rm{rad/s}  

Angular acceleration is the rate of change in angular velocity with respect to time.  

Angular velocity and angular acceleration are related by  

{\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t

Which when re-arranged becomes  

\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}

There’s no change in angular velocity anytime when the angular velocity is \omega = 0\,{\rm{rad/s}}

The equation can be modified as follows:  

\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}

Therefore, the angular acceleration becomes zero hence statement A is valid.  

Statement B  

Angular acceleration is the rate of change in angular velocity with respect to time.  

Angular velocity and angular acceleration are related by  

{\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t

Which when re-arranged becomes  

\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}

There’s no change in angular velocity anytime when the angular velocity is \omega = 10\,{\rm{rad/s}}.The final and initial velocities remain the same.  

The equation can be modified as follows:  

\begin{array}{c}\\\alpha = \frac{{10\,{\rm{rad/s}} - 10\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}

Therefore, the angular acceleration becomes zero and statement B is valid  

Statement C  

Angular velocity is defined as the change in the angular position with respect to time.  

Angular velocity and angular displacement are related by  

\theta = \omega t

Which can also be modified as:  

{\theta _{\rm{f}}} - {\theta _{\rm{i}}}

Note that the final position is {\theta _{\rm{f}}}and initial position is {\theta _{\rm{i}}}

Modifying the equation to find the angular velocity we obtain  

\omega = \frac{{{\theta _{\rm{f}}} - {\theta _{\rm{i}}}}}{t}

When the angular displacement has the same value at all times, the equation becomes  

\begin{array}{c}\\\omega = \frac{{{\theta _{\rm{i}}} - {\theta _{\rm{i}}}}}{t}\\\\ = 0\\\end{array}

The angular velocity becomes zero.  

Angular acceleration and angular velocity are related by  

{\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t

The expression above can be rearranged as follows:  

\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}

At all times, the angular velocity is \omega = 0\,{\rm{rad/s}} hence initial and final velocities remain the same  

We obtain  

\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}

Therefore, the angular acceleration becomes zero and statement C is valid.  

Therefore, statements A,B and C are consistent .

4 0
4 years ago
I finished all my other questions but i don't know how to do this one. could someone help me? I added a lot of points because I
Roman55 [17]

Explanation:

hope it will help you understand, if you have any questions just let me know.

6 0
2 years ago
EASY BRAINLIEST PLEASE HELP!!
jeka94

Answer:

Solution given:

frequency[f]=60,500,000Hz

velocity[V]=300,000,000m/s

wave length=?

we have

wave length=\frac{V}{f}

=\frac{300,000,000}{60,500,000}

=\frac{3000}{605}=4.96 m

Option A.4.96m

3 0
3 years ago
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