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natta225 [31]
3 years ago
12

Suppose that a spring with a larger spring constant was used in this same apparatus. If a given mass were rotated at the same ra

dius at which it had been rotated with the original spring, would the new period of rotation using the new spring be greater or less than the period of rotation using the original spring
Physics
1 answer:
malfutka [58]3 years ago
6 0

Answer:

The new period of rotation using the new spring would be less than the period of rotation using the original spring

Explanation:

Generally the  period of rotation of the mass is mathematically represented as

       T = 2 \pi \sqrt{\frac{I}{k} }

Here I is the moment of inertia of the mass about the rotation axis and  k is the spring constant

Now looking at the equation we can tell that T is inversely proportional to the square root of the spring constant which means that for a larger spring constant the time period would be lesser

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The terminal velocity is not dependent on which one of the following properties? the drag coefficient 1 the force of gravity 2 c
ahrayia [7]
<h2>Answer: the falling time</h2>

Explanation:

When a body or object falls, basically two forces act on it:  

1. The force of air friction, also called<em> </em><u><em>"drag force"</em></u> D:  

D={C}_{d}\frac{\rho V^{2} }{2}A  (1)

Where:  

C_ {d} is the drag coefficient  

\rho is the density  of the fluid (air for example)

V is the velocity  

A is the transversal area of the object

So, this force is proportional to the transversal area of ​​the falling element and to the square of the velocity.  

2. Its <u>weight </u>due to the gravity force W:  

W=m.g

(2)

Where:  

m is the mass of the object

g is the acceleration due gravity  

So, at the moment <u>when the drag force equals the gravity force, the object will have its terminal velocity:</u>

D=W (3)

{C}_{d}\frac{\rho V^{2} }{2}A=m.g  (4)

V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}}  (5) This is the terminal velocity

As we can see, there is no "falling time" in this equation.

Therefore, the terminal velocity is not dependent on the falling time.

6 0
3 years ago
Can someone please help with this problem?
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What causes interstellar dust and clouds to condense in the planets and stars
lorasvet [3.4K]
That would be a nebula, which is an interstellar cloud of hydrogen gas, dust, and plasma. It is the first stage of a star's cycle.
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What human process are carbon sources that are upsetting the balance between CO to update via planets and CO2 released by living
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What is the wavelength of a wave that has a speed of 26 m/s and a frequency of 49? ? Show your work
nirvana33 [79]

Answer:

Wave speed = Wavelength x Frequency

26 m/s = Wavelength x 49

Divide by 49 to find the wavelength:

The wavelength is approximately 0.53

Let me know if this helps!

6 0
3 years ago
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