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SCORPION-xisa [38]
3 years ago
7

Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1

, the spring constant is 2k, the mass is m, and the spring oscillates with amplitude d. In CASE 2, the spring constant is 2k, the mass is 2m, and the spring oscillates with amplitude 2d.
Physics
1 answer:
Artemon [7]3 years ago
4 0

Answer:

\frac{F_1}{F_2} =\frac{1}{2}

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

F_1=2k\times d

case 2:

F_2=2k\times 2d

therefore, \frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}

\frac{F_1}{F_2} =\frac{1}{2}

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Answer:

The position on the x axis is 0.32 m.

Explanation:

Given that,

Point charge = 27 nC

Charge = 6 nC

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We need to calculate the distance

Using formula of electric field

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(r-x)^2}

Put the value into the formula

\dfrac{27\times10^{-9}}{x^2}=\dfrac{6\times10^{-9}}{(1-x)^2}

\dfrac{27}{x^2}=\dfrac{6}{(1-x)^2}

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If the action force is the swimmer pushing water in the leftward direction, what is the reaction force?
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<u>Answer</u><u>:</u>

<u>The </u><u>reaction </u><u>force </u><u>is </u><u>the </u><u>water </u><u>pushing </u><u>the </u><u>swimmer </u><u>in </u><u>the </u><u>rightward </u><u>direction</u><u>.</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

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