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3 years ago

8

3) A person catches a ball with a mass of 145 g dropped from a height of 60.0 m above his glove. His hand stops the ball in 0.01

00 s. What is the force exerted by his glove on the ball? Assume the ball slows down with constant acceleration.

1 answer:

iragen [17]3 years ago

7 0

Answer: 87KN

Explanation: F= m(v-u)/t

V= h/t

V= 60/0.0100

v= 6000m/s

M=145g/1000=0.145kg

F= 0.145*6000/0 .001

F= 87000N

F= 87KN

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3 years ago

in a solar system far, far away the sun's intensity is 200 w/m2 for an inner planet located a distance r away. what is the sun's

GarryVolchara [31]

The sun's intensity for an outer planet located at a distance 6r from the sun is 5.55 W/m². The result is obtained by using the inverse square law formula.

<h3>What is the Inverse Square Law formula?</h3>

The Inverse Square Law formula describes the intensity of light is inversely proportional to the square of the distance. It can be expressed as

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

Where

I₁ = Intensity at distance 1 (W/m²)
I₂ = Intensity at distance 2 (W/m²)
d₁ = distance 1 from a light source (m)
d₂ = distance 2 from a light source (m)

Given the case the sun's intensity is 200 W/m² for an inner planet at the distance r. If an outer planet is at a distance 6r, what is the sun's intensity?

By using the inverse square law formula, the sun's intensity for an outer planet is

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

$\frac{200 }{I_{2} } = \frac{(6r)^{2} }{r^{2}}$

$\frac{200 }{I_{2} } = \frac{36r^{2} }{r^{2}}$

$I_{2} = \frac{200} {36}$

I₂ = 5.55 W/m²

Hence, the sun's intensity for a planet at a distance 6r from the sun is 5.55 W/m².

Learn more about intensity of light here:

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3 0

1 year ago

A 2.66 kg, 22.25 cm diameter turntable rotates at 238 rpm on frictionless bearings. Two 420 g blocks fall from above, hit the tu

Phantasy [73]

Answer:

The value is $w_f = 146 \ rpm$

Explanation:

From the question we are told that

The mass is $m = 2.66 \ kg$

The diameter is $d = 22.25 \ cm = 0.2225 \ m$

The angular speed is $w_i = 238 \ rpm$

The mass of each of the blocks is $m_b = 420 \ g = 0.420 \ kg$

Generally the radius of the turntable is mathematically represented as

$r = \frac{d}{2}$

=> $r = \frac{0.2225}{2}$

=> $r = 0.11125$

The moment of inertia of the turntable before the blocks fell is mathematically represented as

$I_{i} = \frac{1}{2} * m * r^2$

=> $I_{i} = \frac{1}{2} * 2.66 * (0.11125)^2$

=> $I_{i} = 0.01646 \ kg\cdot m^2$

The moment of inertia of the turntable after the blocks fell is mathematically represented as

$I_{f} = \frac{1}{2} * m * r^2 + 2 m_b * r^2$

=> $I_{f} = \frac{1}{2} * 2.66 * (0.11125)^2 + 2 * 0.420 * 0.11125^2$

=> $I_{f} = 0.02686 \ kg\cdot m^2$

Generally from the law of angular momentum conservation

$L_i = L_f$

Here $L_i$ is the initial angular momentum of the turntable before the blocks fell which is mathematically represented as

$L_i = I_i * w_i$

and $L_f$ is the initial angular momentum of the turntable after the blocks fell which is mathematically represented as

$L_i = I_f * w_f$

So

$0.01646* 238 = 0.02686 * w_f$

=> $w_f = 146 \ rpm$

5 0

2 years ago