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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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The weight 7.61x10^2N

The angle 26.0

To determine the magnitude of horizontal force applied

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F_t=7.61x10^2N*tan(26.0)

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Explanation:

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Write the mathemetical relation between work force and displacement​
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A sled is accelerating down a hill at a rate of 1 m s2 . If the mass of the sled is suddenly cut in half and the net force on th
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We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
a'= \frac{2F}{ \frac{m}{2} } = \frac{4F}{m} = 4\frac{F}{m}
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4 0
3 years ago
Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.4 cmcm and a current of 12 AA. The bigger
Free_Kalibri [48]

Answer:

Explanation:

Given that,

Current in loops are

i1 = 12A

i2 = 20A

The loops are 3.4cm apart

The magnetic field at the center is found to be zero, so when want to find the radius of bigger loop

Magnetic Field is given as

B= μoi/2πr

Where,

μo is a constant = 4π×10^-7 Tm/A

r is the distance between the two wires

i is the current in the wires

B is the magnetic field

NOTE

Field due to large loop should be equal to the smaller loop.

B1 = B2

μo•i1 / 2π•r1 = μo•i2 / 2π•r2

Then, μo, 2π cancels out, so we have

i1 / r1 = i2 / r2

Make r2 subject of formula

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r2 = i2•r1 / i2

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r2 = 5.67cm

The radius of the bigger loop is 5.67cm.

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