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Snowcat [4.5K]
1 year ago
5

what happens to the frequency the centripetal force on an object in circular motion is doubled while the object keeps the same r

adius of rotation
Physics
1 answer:
kondaur [170]1 year ago
3 0

The new frequency also double the centripetal force on an object in circular motion is doubled while the object keeps the same radius of rotation.

The mass of the object is m .

The radius of the circle is r.

The new centripetal force F2 = 2F1 .

A force that causes a body to follow a curved path is known as a centripetal force, which derives from the Latin words centrum, "Centre," and petered, "to pursue." It always moves in a direction that is the opposite of the body's motion and in the direction of the fixed point of the path's immediate Centre of curvature. "A force by which bodies are dragged or driven, or in any other manner tend, towards a point as to a Centre," was how Isaac Newton defined it. The centripetal force that causes astronomical orbits in Newtonian physics is provided by gravity.

The scenario in which a body moves uniformly fast along a circular path is one frequent example of centripetal force. Towards the Centre of the object, the centripetal force is directed along the radius and at an angle to the velocity.

Learn more about centripetal force  here:

brainly.com/question/11324711

#SPJ4

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Could a car drive on a frictionless surface? Explain using the terms action
Julli [10]

Answer:

No, it cannot. The car needs the friction of the surface to drive because the car pushes the surface backwards, and the surfaces makes a reaction force pushing the car forward, and that works because of the friction. In a frictionless surface the tires would rotate in the same place

7 0
3 years ago
Clasifica los siguientes enunciados de acuerdo a si se describen movimientos
Alenkasestr [34]

Answer:

- translation

- rotation, traslation

- traslation, rotation

- vibrating

Explanation:

El movimiento de un cuerpo cae por su propio peso <u>traslación</u>.

El movimiento de las ruedas de una bicicleta al ser pedaleada <u>rotación, traslación</u>.

El movimiento de la Tierra alrededor de sol <u>traslación, rotación</u>.

El movimiento de la cuerda de una guitarra cuando se está tocando música <u>vibración</u>.​

- - - - - - - - - - - - - - - - - - - -  - - - - - - - -  

The movement of a body falls under its own  weight <u>translation</u>.

The movement of the wheels of a bicycle when being pedaled <u>rotation, translation.</u>

The movement of the Earth around the sun, <u>translation, rotation</u>.

The movement of a guitar string when playing  music <u>vibrating</u>.

7 0
3 years ago
The mass of the earth is 5.96/10kg^24. The radius of the earth is approximately 6.37x10^6 calculate the force of gravity
n200080 [17]
<h2>Answer:g=9.79ms^{-2},A object of mass m at the surface of earth experiences a force mg</h2>

Explanation:

Let M be the mass of earth.

Let R be the radius of earth.

Let G be the universal gravitational constant.

Given,

M=5.96\times 10^{24}Kg

R=6.37\times 10^{6}m

G=6.67259 \times 10^{-11}Nm^{2}Kg^{-2}

Let g be the acceleration due to gravity.

Then,g=\dfrac{GM}{R^{2}}

g=\frac{6.67259 \times 10^{-11}\times 5.96\times 10^{24}}{(6.37\times 10^{6})^{2}}

g=9.79ms^{-2}

A object of mass m at the surface of earth experiences a force mg

3 0
3 years ago
N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e
Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

KE_i+PE_i=KE_f+PE_f

Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)

from the law of conservation of the linear momentum

m_1v_1=m_2v_2

Therefore,

Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)

=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]

v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]

Substitute the values in the above result

v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]

=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s

B)  the speed of the sphere with mass 107.0 kg is

v_2=\frac{m_1v_1}{m_2}

=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s

C)  the magnitude of the relative velocity with which one sphere is

v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s

D) the distance of the centre is proportional to the acceleration

\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962

Thus,

x_1=3.962x_2

and

x_2=0.252x_1

When the sphere make contact with eachother

Therefore,

x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r

And

x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r

The point of contact of the sphere is

32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m

3 0
3 years ago
4. How does the type of medium affect a sound wave?
jek_recluse [69]

Answer:

The type of medium affects a sound wave as sound travels with the help of the vibration in particles.

Explanation:

As different mediums have different amount and size of particles, for example, the speed of sound is faster through solid than liquid as solids have closely packed particles whereas liquids are loosely packed.

5 0
3 years ago
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