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krek1111 [17]
2 years ago
10

A 20-kg boy slides down a smooth, snow-covered hill on a plastic disk. The hill is at a 10° angle to the horizontal, and the slo

pe is 50 m long. Part A If the boy starts from rest, what is his speed at the bottom of the hill?​
Physics
1 answer:
zvonat [6]2 years ago
7 0

Answer:

13 m/s

Explanation:

I assume we are ignoring friction.

The boy's PE will all be converted to KE at the bottom of the hill.

to find PE = mgh   we need to know h

   h = 50 sin 10 = 8.68 meters

     then:    PE = 20 * 9.81 * 8.68 =<u> 1703.49</u> j

KE = 1/2 m v^2 = <u>1703 .49</u>

            v = 13 m/s

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If they become closer, it is increased, and if the objects become farther away is decreased.
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3 years ago
A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis t
Likurg_2 [28]

Answer:

8.91 J

Explanation:

mass, m = 8.20 kg

radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

                                                    = 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2

n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

K = \frac{1}{2}I\omega ^{2}

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

6 0
3 years ago
Calculate the density of a tin of mass 100g whose dimensions are 2cmx5cmx​
Luda [366]

Answer:

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7 0
3 years ago
Two resistors, one with 7.00 Ω of resistance and the other with 11.00 Ω of resistance, are connected in series to a 9.00 V batte
IRINA_888 [86]

Answer:

4.5 W

Explanation:

Applying,

P = V²/(R₁+R₂).................. Equation 1

Where P = Power, V = Voltage, R₁ and R₂ = values of the two resistor.

From the question,

Given: V = 9.00 V, R₁ = 7.00 Ω, R₂ = 11.00  Ω

Substitute these values into equation 1

P = 9²/(7+11)

P = 81/(18)

P = 4.5 Watt.

Hence the power dessipated by the two resistors is 4.5 watt

5 0
3 years ago
What eccentricity value results in a circular orbit?
Oxana [17]

Answer:

Zero

Explanation:

Given the equation of an ellipse:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

The eccentrity of an ellipse is given by:

e=\sqrt{1-\frac{b^2}{a^2}}

For a circle, we have

a=b

Therefore the eccentricity of a circle is

e=\sqrt{1-\frac{1^2}{1^2}}=0

7 0
3 years ago
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