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Dennis_Churaev [7]
3 years ago
6

recall that a centripetal force is not a new type of force: it's a role that is taken by one of the four types of force that we

already use. if we see an object moving in a curved path, we know that at least one of those forces is pointing inward, acting centripetally. describe at least one real-life example of an object moving in a curved path where its centripetal force is provided at least partially by each of the following types of forces: gravity, normal force, tension,
Physics
1 answer:
Firdavs [7]3 years ago
7 0

Answer:

Answer in explanation.

Explanation:

<u>GRAVITY AS CENTRIPETAL FORCE</u>:

When a satellite orbits around a planet, it is in a circular motion. In this scenario, the centripetal force is provided by the gravity force.

<u>NORMAL FORCE AS CENTRIPETAL FORCE</u>:

When a spacecraft is sent into space, it rotates about its own axis to create artificial gravity. During this rotating motion, the centripetal force is provided by the normal reaction (force) of the walls of the satellite.

<u>TENSION AS CENTRIPETAL FORCE</u>:

When a mass attached to a rope is swirled in a circular motion, the tension in rope acts as the centripetal force.  

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Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the
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KE_A=33\ J

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Now when the compression of the particles with the spring is released, the spring potential energy must get converted into the kinetic energy of the particles and their momentum must be conserved.

Kinetic energy:

\frac{1}{2}m_A.v_A^2+\frac{1}{2}m_B.v_B^2=132

3m.v_A^2+m.v_B^2=264 .............................(1)

<u>Using the conservation of linear momentum:</u>

m_A.v_A+m_B.v_B=0

3m.v_A+m.v_B=0 .............................(2)

Put the value of v_A from eq. (2) into eq. (1)

3m\times (\frac{-v_B}{3})^2+m.v_B^2=264

v_B^2=\frac{198}{m}  ...........................(3)

<u>Now the kinetic energy of particle B:</u>

KE_B=\frac{1}{2}\times m_B\times v_B^2

KE_B=\frac{1}{2}\times m\times \frac{198}{m}

KE_B=99\ J

Put the value of v_B^2 form eq. (3) into eq. (1):

v_A^2=\frac{22}{m}

<u>Now the kinetic energy of particle A:</u>

<u />KE_A=\frac{1}{2}m_A.v_A^2<u />

<u />KE_B=\frac{1}{2}\times 3m\times \frac{22}{m}<u />

KE_A=33\ J

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3 years ago
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