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igomit [66]
3 years ago
15

Match the object to the type of motion it's displaying. which letter goes with which number?

Physics
1 answer:
makkiz [27]3 years ago
8 0

<em>Hello There!!</em>

<em>I think the best answer choice is:</em>

<em>A. </em>Linear motion

<em>B. </em>Rotational motion

<em>D. </em>Vibrational motion

<em>4. </em>Runners sprinting across a straight track

<em>C. </em>Curvilinear motion

<em>3. </em>Music being\\\\\\ played on a stringed instrument, such as a guitar

<em>2. </em>A car driving up a winding mountain road

<em>1. </em>The propeller of an aircraft in midair

<em>I'm not sure this is 100% right....</em>

<em>P.S </em><em>Tell me if this is wrong....</em>

<em />Hope This Helps<em> </em>Sangwoo!!<em />

<em>#</em>Be<em> </em>Bold<em />

<em># </em>Always<em> </em>Brainly<em />

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7 0
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If a series circuit contains a 12-V battery, a 6-ohm resistor, and a 4-ohm resistor, what is the current in the circuit?
Shalnov [3]

In a series circuit the total current is the same throughout resistors and so:

I_{total}=I_1=I_2

The voltage is distributed throughout the resistors and so:

V_{total}=V_1+V_2

and the total resistance can be calculated by adding up the resistors resistance:

R_{total}=R_1+R_2

First thing is to calculate the total resistance and so:

R_{total}=6\Omega + 4\Omega = 10\Omega

And by Omh's law V=IR we have:

V_{total}=I_{total}R_{total}\\\\I_{total}=\frac{V_{total}}{R_{total}}= \frac{12V}{10\Omega} =1.2A

And so the total current of the circuit is 1.2 amps i.e. 1.2 A.


6 0
3 years ago
Read 2 more answers
A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h
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Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

            \alpha_{1} = \frac{1}{4} \alpha_{2}

Thus, we can conclude that the new rotation is \frac{1}{4} times that of the first rotation rate.

8 0
3 years ago
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