Math / Algebra

Question(I) A horizontal force of 310 N is exerted on a $2.0-\mathrm{kg}$ ball as it rotates (at arm's length) uniformly in a horizontal circle of radius 0.90 m . Calculate the speed of the ball.

Studdy Solution

STEP 1

1. The ball is rotating uniformly in a horizontal circle.
2. The radius of the circle is 0.90 m.
3. The force exerted on the ball is 310 N.
4. The mass of the ball is 2.0 kg.
5. We are trying to find the speed of the ball.

STEP 2

1. Understand the relationship between force, mass, and centripetal acceleration.
2. Use the formula for centripetal force to find the speed of the ball.

STEP 3

Understand the relationship between force, mass, and centripetal acceleration.
The centripetal force $F_c$ required to keep an object moving in a circle of radius $r$ at speed $v$ is given by the formula:
$F_c = \frac{mv^2}{r}$
where: - $F_c$ is the centripetal force, - $m$ is the mass of the object, - $v$ is the speed of the object, - $r$ is the radius of the circle.

STEP 4

Use the formula for centripetal force to find the speed of the ball.
Given: - $F_c = 310 \, \text{N}$ - $m = 2.0 \, \text{kg}$ - $r = 0.90 \, \text{m}$
Substitute these values into the centripetal force formula:
$310 = \frac{2.0 \times v^2}{0.90}$

STEP 5

Solve for $v^2$ .
Multiply both sides by 0.90 to isolate $v^2$ :
$310 \times 0.90 = 2.0 \times v^2$
$279 = 2.0 \times v^2$

STEP 6

Divide both sides by 2.0 to solve for $v^2$ :
$v^2 = \frac{279}{2.0}$
$v^2 = 139.5$

STEP 7

Take the square root of both sides to solve for $v$ :
$v = \sqrt{139.5}$
$v \approx 11.81 \, \text{m/s}$
The speed of the ball is approximately:
$\boxed{11.81 \, \text{m/s}}$

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Question(I) A horizontal force of 310 N is exerted on a $2.0-\mathrm{kg}$ ball as it rotates (at arm's length) uniformly in a horizontal circle of radius 0.90 m . Calculate the speed of the ball.

Studdy Solution

STEP 1

1. The ball is rotating uniformly in a horizontal circle.
2. The radius of the circle is 0.90 m.
3. The force exerted on the ball is 310 N.
4. The mass of the ball is 2.0 kg.
5. We are trying to find the speed of the ball.

STEP 2

1. Understand the relationship between force, mass, and centripetal acceleration.
2. Use the formula for centripetal force to find the speed of the ball.

STEP 3

STEP 4

Use the formula for centripetal force to find the speed of the ball.
Given: - $F_c = 310 \, \text{N}$ - $m = 2.0 \, \text{kg}$ - $r = 0.90 \, \text{m}$
Substitute these values into the centripetal force formula:
$310 = \frac{2.0 \times v^2}{0.90}$

STEP 5

Solve for $v^2$ .
Multiply both sides by 0.90 to isolate $v^2$ :
$310 \times 0.90 = 2.0 \times v^2$
$279 = 2.0 \times v^2$

STEP 6

Divide both sides by 2.0 to solve for $v^2$ :
$v^2 = \frac{279}{2.0}$
$v^2 = 139.5$

STEP 7

Take the square root of both sides to solve for $v$ :
$v = \sqrt{139.5}$
$v \approx 11.81 \, \text{m/s}$
The speed of the ball is approximately:
$\boxed{11.81 \, \text{m/s}}$

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Question(I) A horizontal force of 310 N is exerted on a 2.0−kg2.0-\mathrm{kg}2.0−kg ball as it rotates (at arm's length) uniformly in a horizontal circle of radius 0.90 m . Calculate the speed of the ball.

Studdy Solution

STEP 1

1. The ball is rotating uniformly in a horizontal circle.2. The radius of the circle is 0.90 m.3. The force exerted on the ball is 310 N.4. The mass of the ball is 2.0 kg.5. We are trying to find the speed of the ball.

STEP 2

1. Understand the relationship between force, mass, and centripetal acceleration.2. Use the formula for centripetal force to find the speed of the ball.

STEP 3

STEP 4

STEP 5

Solve for v2 v^2 v2.Multiply both sides by 0.90 to isolate v2 v^2 v2:310×0.90=2.0×v2 310 \times 0.90 = 2.0 \times v^2 310×0.90=2.0×v2279=2.0×v2 279 = 2.0 \times v^2 279=2.0×v2

STEP 6

Divide both sides by 2.0 to solve for v2 v^2 v2:v2=2792.0 v^2 = \frac{279}{2.0} v2=2.0279​v2=139.5 v^2 = 139.5 v2=139.5

STEP 7

Take the square root of both sides to solve for v v v:v=139.5 v = \sqrt{139.5} v=139.5​v≈11.81 m/s v \approx 11.81 \, \text{m/s} v≈11.81m/sThe speed of the ball is approximately:11.81 m/s \boxed{11.81 \, \text{m/s}} 11.81m/s​

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question(I) A horizontal force of 310 N is exerted on a $2.0-\mathrm{kg}$ ball as it rotates (at arm's length) uniformly in a horizontal circle of radius 0.90 m . Calculate the speed of the ball.

1. The ball is rotating uniformly in a horizontal circle.
2. The radius of the circle is 0.90 m.
3. The force exerted on the ball is 310 N.
4. The mass of the ball is 2.0 kg.
5. We are trying to find the speed of the ball.

1. Understand the relationship between force, mass, and centripetal acceleration.
2. Use the formula for centripetal force to find the speed of the ball.

Solve for $v^2$ .
Multiply both sides by 0.90 to isolate $v^2$ :
$310 \times 0.90 = 2.0 \times v^2$
$279 = 2.0 \times v^2$

Divide both sides by 2.0 to solve for $v^2$ :
$v^2 = \frac{279}{2.0}$
$v^2 = 139.5$

Take the square root of both sides to solve for $v$ :
$v = \sqrt{139.5}$
$v \approx 11.81 \, \text{m/s}$
The speed of the ball is approximately:
$\boxed{11.81 \, \text{m/s}}$