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4 years ago

Which force keeps an object moving in a circle

Physics

2 answers:

AlekseyPX4 years ago

8 0

Answer:

Centripetal force

Explanation:

An object moving in a circle experiences a centripetal acceleration. The force that causes this acceleration is called centripetal force.

NeX [460]4 years ago

7 0

Answer:

The force is called Centripetal Force

Explanation:

Force simply means a push or pull on an object which causes it to move from it's original position.

We have different types of force, one of which is the Centripetal Force

The Centripetal Force is a force that makes a body to move in an circular position. Example of Centripetal Force is turning a car.

Centripetal Force is normally calculated with the formula

Fc = mv²/r

Where:

Fc = centripetal force

m = mass

v = velocity

r = radius

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3 years ago

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A penny is placed on a rotating turntable. where on the turntable does the penny require the largest centripetal force to remain

Mama L [17]

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

w = angular speed of rotation of turntable

F = centripetal force experienced by the penny

centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as

F = m r w²

in the above equation , mass of penny "m" and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

hence greater the distance from center , greater will be the centripetal force to remain in place.

So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.

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3 years ago

A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i

Darya [45]

The position of the first ball is

$y_1=h-\dfrac g2t^2$

while the position of the second ball, thrown with initial velocity $v$ , is

$y_2=vt-\dfrac g2t^2$

The time it takes for the first ball to reach the halfway point satisfies

$\dfrac h2=h-\dfrac g2t^2$

$\implies\dfrac h2=\dfrac g2t^2$

$\implies t=\sqrt{\dfrac hg}$

We want the second ball to reach the same height at the same time, so that

$\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2$

$\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)$

$\implies h=v\sqrt{\dfrac hg}$

$\implies v=\sqrt{hg}$

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3 years ago

Projectile Motion—A tennis ball is thrown out a window 28 m above the ground at an initial velocity of 15.0 m/s and 20.0° below

NNADVOKAT [17]

Answer:

The distance will be x = 41.7 [m]

Explanation:

We must first find the components in the x & y axes of the initial velocity.

$(v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s]$

The acceleration is the gravity acceleration therefore.

g = 9.81 [m/s^2]

Now we can calculate how long it takes to fall.

$y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s]$

With this time we can find the horizontal distance that runs the projectile.

$x=(v_{o})_{x}*t\\x=14.09*2.96\\x=41.7[m]$

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3 years ago

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Answer:

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Explanation:

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3 years ago