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Shalnov [3]
4 years ago
8

When the displacement of a mass on a spring is 12 a the half of the amplitude, what fraction of the mechanical energy is kinetic

energy?
Physics
1 answer:
Sveta_85 [38]4 years ago
7 0
You know that when the displacement is equal to the amplitude (A), the velocity is zero, which implies that the kinetic energy (KE) is zeero, so the total mechanical energy (ME) is the potential energy (PE).

And you know that the potential energy, PE, is [ 1/2 ] k (x^2)

Then, use x = A, to calculate the PE in the point where ME = PE.

ME = PE = [1/2] k (A)^2.

At half of the amplitude, x = A/2 => PE = [ 1/2] k (A/2)^2

=> PE = [1/4] { [1/2]k(A)^2 } = .[1/4] ME

So, if PE is 1/4 of ME, KE is 3/4 of ME.

And the answer is 3/4


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Two tiny, spherical water drops, with identical charges of −8.00 ✕ 10^(−17) C, have a center-to-center separation of 2.00 cm.
Damm [24]

Answer:

F=1.4384×10⁻¹⁹N

Explanation:

Given Data

Charge q= -8.00×10⁻¹⁷C

Distance r=2.00 cm=0.02 m

To find

Electrostatic force

Solution

The electrostatic force between between them can be calculated from Coulombs law as

F=\frac{kq^{2} }{r^{2} }

Substitute the given values we get

F=\frac{(8.99*10^{9} )*(-8.00*10^{-17} )^{2} }{(0.02)^{2} }\\ F=1.4384*10^{-19} N

7 0
3 years ago
A 2500-N net force acting on a 880-kg car accelerates it at a rate of ______ m/s/s
Bond [772]

Answer:

a = 2.84 m/s²

Explanation:

Given that,

Net force, F = 2500 N

Mass of the car, m = 880 kg

We need to find the acceleration of the car. Net force is given by :

F = ma

a=\dfrac{F}{m}\\\\a=\dfrac{2500\ N}{880\ m/s^2}\\\\a=2.84\ m/s^2

So, the acceleration of the car is 2.84 m/s².

8 0
3 years ago
Diffraction supports the:<br><br> A. wave theory of light.<br><br> B. particle theory of light.
Artemon [7]
The answer is a hope its helps you
8 0
3 years ago
Read 2 more answers
What is the speed of a sound wave that takes 0.5 s to travel 750 m?
kvasek [131]

Answer:

C. 1500.

Explanation:

750 / .5 = 1500.

Hope this helps & best of luck!

Feel free to message me if you need more help! :)

4 0
3 years ago
A 0.0663 kg ingot of metal is heated to 241◦C
Westkost [7]

Answer:280.216j/kg°C

Explanation:

Mass of metal=0.0663kg

mass of water=0.395kg

Final temperature=27.4°C

Temperature of metal=241°C

Temperature of water=25°C

specific heat capacity of water=4186j/kg°C

0.0663xax(241-27.4)=0.395x4186x(27.4-25)

0.0663xax213.6=0.395x4186x2.4

14.16168a=3968.328

a=3968.328 ➗ 14.16168

a=280.216j/kg°C

4 0
3 years ago
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