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Georgia [21]
4 years ago
13

The net torce on an object moving with constant speed in circular motion is in which direction?

Physics
1 answer:
aleksley [76]4 years ago
6 0

The correct answer is C) towards the center of the circle.

Although the object is moving at a constant speed it is constantly accelerating due to the constant change in direction as it describes the circular path. This causes a constant change in velocity as velocity is a vector quantity.

For the object to maintain the circular path there has to be centripetal force acting on the object and this centripetal force is directed towards the center of the circle.

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Answer:

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Hope this helps

7 0
3 years ago
The jogger ran 3km east.
erma4kov [3.2K]

Answer: d

Explanation:

7 0
4 years ago
A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that
Marat540 [252]

Answer:

Part(a): The value of the spring constant is 3.11 \times 10^{2}~Kg~s^{-2}.

Part(b): The work done by the variable force that stretches the collagen is 1.5 \times 10^{-6}~J.

Explanation:

Part(a):

If 'k' be the force constant and if due the application of a force 'F' on the collagen '\Delta l' be it's increase in length, then from Hook's law

F = k~\Delta l....................................................................(I)

Also, Young's modulus of a material is given by

Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)

where 'A' is the area of the material and 'l' is the length.

Comparing equation (I) and (II) we can write

&& Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}

Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.

Substituting the given values in equation (III), we have

k = \dfrac{3.10 \times 10^{6}~N~m^{-2} \times \pi \times (0.00093)^{2}~m^{2}}{.027~m} = 3.11 \times 10^{2}~Kg~s^{-2}

Part(b):

We know the amount of work done (W) on the collagen is stored as a potential energy (U) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as

W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)

Substituting all the values, we can write

W = \dfrac{(3.06 \times 10^{-2})^{2}~N^{2}}{2 \times 3.11 \times 10^{2}~Kg~s^{-2}} = 1.5 \times 10^{-6}~J

3 0
3 years ago
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Ivanshal [37]

If the wavelength increases (gets longer), then the frequency <em>decreases</em>.

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3 years ago
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AveGali [126]
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