Math / Algebra

Question40. (II) A $0.45-\mathrm{kg}$ ball, attached to the end of a horizontal cord, is revolved in a circle of radius 1.3 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 75 N , what is the maximum speed the ball can have?

Studdy Solution

STEP 1

1. The mass of the ball is $0.45 \, \text{kg}$ .
2. The radius of the circle is $1.3 \, \text{m}$ .
3. The maximum tension the cord can withstand is $75 \, \text{N}$ .
4. The surface is frictionless.
5. We are trying to find the maximum speed of the ball.

STEP 2

1. Understand the relationship between tension, mass, radius, and speed.
2. Write the formula for centripetal force.
3. Substitute known values into the formula.
4. Solve for the maximum speed.

STEP 3

Understand the relationship between tension, mass, radius, and speed.
The tension in the cord provides the centripetal force necessary to keep the ball moving in a circle. The formula for centripetal force $F_c$ is:
$F_c = \frac{mv^2}{r}$
where $m$ is the mass, $v$ is the speed, and $r$ is the radius.

STEP 4

Write the formula for centripetal force.
Since the tension in the cord is the centripetal force, we have:
$T = \frac{mv^2}{r}$
where $T$ is the tension.

STEP 5

Substitute known values into the formula.
Given $T = 75 \, \text{N}$ , $m = 0.45 \, \text{kg}$ , and $r = 1.3 \, \text{m}$ , substitute these into the equation:
$75 = \frac{0.45 \times v^2}{1.3}$

STEP 6

Solve for the maximum speed.
First, isolate $v^2$ by multiplying both sides by $1.3$ :
$75 \times 1.3 = 0.45 \times v^2$
$97.5 = 0.45 \times v^2$
Next, divide both sides by $0.45$ :
$v^2 = \frac{97.5}{0.45}$
$v^2 = 216.67$
Finally, take the square root of both sides to solve for $v$ :
$v = \sqrt{216.67}$
$v \approx 14.72 \, \text{m/s}$
The maximum speed the ball can have is approximately:
$\boxed{14.72 \, \text{m/s}}$

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Question40. (II) A $0.45-\mathrm{kg}$ ball, attached to the end of a horizontal cord, is revolved in a circle of radius 1.3 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 75 N , what is the maximum speed the ball can have?

Studdy Solution

STEP 1

1. The mass of the ball is $0.45 \, \text{kg}$ .
2. The radius of the circle is $1.3 \, \text{m}$ .
3. The maximum tension the cord can withstand is $75 \, \text{N}$ .
4. The surface is frictionless.
5. We are trying to find the maximum speed of the ball.

STEP 2

1. Understand the relationship between tension, mass, radius, and speed.
2. Write the formula for centripetal force.
3. Substitute known values into the formula.
4. Solve for the maximum speed.

STEP 3

STEP 4

Write the formula for centripetal force.
Since the tension in the cord is the centripetal force, we have:
$T = \frac{mv^2}{r}$
where $T$ is the tension.

STEP 5

Substitute known values into the formula.
Given $T = 75 \, \text{N}$ , $m = 0.45 \, \text{kg}$ , and $r = 1.3 \, \text{m}$ , substitute these into the equation:
$75 = \frac{0.45 \times v^2}{1.3}$

STEP 6

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Question40. (II) A 0.45−kg0.45-\mathrm{kg}0.45−kg ball, attached to the end of a horizontal cord, is revolved in a circle of radius 1.3 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 75 N , what is the maximum speed the ball can have?

Studdy Solution

STEP 1

1. The mass of the ball is 0.45 kg 0.45 \, \text{kg} 0.45kg.2. The radius of the circle is 1.3 m 1.3 \, \text{m} 1.3m.3. The maximum tension the cord can withstand is 75 N 75 \, \text{N} 75N.4. The surface is frictionless.5. We are trying to find the maximum speed of the ball.

STEP 2

1. Understand the relationship between tension, mass, radius, and speed.2. Write the formula for centripetal force.3. Substitute known values into the formula.4. Solve for the maximum speed.

STEP 3

STEP 4

Write the formula for centripetal force.Since the tension in the cord is the centripetal force, we have:T=mv2r T = \frac{mv^2}{r} T=rmv2​where T T T is the tension.

STEP 5

Substitute known values into the formula.Given T=75 N T = 75 \, \text{N} T=75N, m=0.45 kg m = 0.45 \, \text{kg} m=0.45kg, and r=1.3 m r = 1.3 \, \text{m} r=1.3m, substitute these into the equation:75=0.45×v21.3 75 = \frac{0.45 \times v^2}{1.3} 75=1.30.45×v2​

STEP 6

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Question40. (II) A $0.45-\mathrm{kg}$ ball, attached to the end of a horizontal cord, is revolved in a circle of radius 1.3 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 75 N , what is the maximum speed the ball can have?

1. The mass of the ball is $0.45 \, \text{kg}$ .
2. The radius of the circle is $1.3 \, \text{m}$ .
3. The maximum tension the cord can withstand is $75 \, \text{N}$ .
4. The surface is frictionless.
5. We are trying to find the maximum speed of the ball.

1. Understand the relationship between tension, mass, radius, and speed.
2. Write the formula for centripetal force.
3. Substitute known values into the formula.
4. Solve for the maximum speed.

Write the formula for centripetal force.
Since the tension in the cord is the centripetal force, we have:
$T = \frac{mv^2}{r}$
where $T$ is the tension.

Substitute known values into the formula.
Given $T = 75 \, \text{N}$ , $m = 0.45 \, \text{kg}$ , and $r = 1.3 \, \text{m}$ , substitute these into the equation:
$75 = \frac{0.45 \times v^2}{1.3}$