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3 years ago

A centripetal force causes circular motion because it accelerates an object

Physics

2 answers:

MAVERICK [17]3 years ago

6 0

The answer is option A.

Centripetal force is always directed towards the centre and does not change the speed of the body,but there is a change in the direction.

ElenaW [278]3 years ago

3 0

Answer:

Towards the center of circle, which changes the direction of motion but not speed

Explanation:

The force acting on an object when it is moving in circular path is called centripetal force. The formula for centripetal force is given by :

$F=\dfrac{mv^2}{r}$

Where

m = mass

v = velocity

r = radius of circle

In a circular path, the velocity of object changes continuously while its speed remains constant. The centripetal force acts towards the center of circle. Hence, the correct option is (a) " towards the center of circle, which changes the direction of motion but not speed ".

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A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find

Ludmilka [50]

Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

m = mass of the ladder, 7.98 kg

L = length of the ladder, 4.15 m

On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

I = 137.44 / 12

I = 11.45 kg·m²

That is the moment of inertia about the center.

Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that

τ = 11.453 kg·m² * 0.395 rad/s²

τ = 4.535 N·m

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2 years ago

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2 years ago

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00

vivado [14]

Answer:

7.5 m

Explanation:

$v$ = initial speed of the ball = 8 m/s

$\theta$ = angle of launch = 40° deg

Consider the motion along the vertical direction :

$v_{oy}$ = initial velocity along vertical direction = $v Sin\theta$ = 8 Sin40 = 5.14 m/s

$a_{y}$ = acceleration along vertical direction = - 9.8 m/s²

$t$ = time of travel

$y$ = vertical displacement = - 1 m

Using the kinematics equation

$y = v_{oy}t + (0.5)a_{y}t^{2}$

$- 1 = (5.14)t + (0.5)(- 9.8)t^{2$

$t$ = 1.22 sec

Consider the motion along the horizontal direction :

$v_{ox}$ = initial velocity along horizontal direction = $v Sin\theta$ = 8 Cos40 = 6.13 m/s

$a_{x}$ = acceleration along vertical direction = 0 m/s²

$t$ = time of travel = 1.22 sec

$x$ = horizontal displacement = ?

Using the kinematics equation

$x = v_{ox}t + (0.5)a_{x}t^{2}$

$x = (6.13)(1.22) + (0.5)(0)(1.22)^{2$

$x$ = 7.5 m

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2 years ago

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Aneli [31]

Answer:

The forces are of the same magnitude just opposite directions

Explanation:

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sasho [114]

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