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Sladkaya [172]
3 years ago
13

A 120 g block attached to a spring with spring constant 3.0 N/m oscillates horizontally on a frictionless table. Its velocity is

21 cm/s when x0 = -4.1 cm
Physics
1 answer:
Marysya12 [62]3 years ago
4 0

Answer:

0.059 m

Explanation:

0.5k A^2 = 0.5 (k x^{2} + m v^{2}) where k is the spring constant, A is the amplitude, x is the extension of spring, m is the mass of the spring and v is the velocity in m/s. Making a the subject then

A = \sqrt{(x^{2} + (\frac {mv^{2}}{k})}

By substituting the given values then where m is 0.12 Kg, x= 0.0041 m, v is 0.21 m/s and K is 3 N/m then

A = \sqrt{(0.0041^{2} + (\frac {0.12\times 0.21^{2}}{3})}=0.058694122m\approx 0.059 m

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A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work
kifflom [539]

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Explanation:

Work done is the product of applied force and displacement of the object in the direction of force.

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3 0
3 years ago
If a person weighs 818 N on earth and 5320 N on the surface of a nearby planet, what is the acceleration due to gravity on that
alexandr402 [8]

Answer:

g'=63.74\ m/s^2

Explanation:

It is given that,

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Weight of a person is given by the following formula as :

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m=\dfrac{818\ N}{9.8\ m/s^2}

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The mass of an object is same everywhere. It does not depend on the location.

Let W' is the weight of the person on the surface of a nearby planet, W' = 5320 N

g' is the acceleration due to gravity on that planet. So,

g'=\dfrac{W'}{m}

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So, the acceleration due to gravity on that planet is 63.74\ m/s^2. Hence, this is the required solution.                                                                    

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