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Charra [1.4K]
3 years ago
14

You pull your little sister across a flat snowy field on a sled. Your sister plus the sled have a mass of 22 kg. The rope is at

an angle of 40 degrees to the ground. As you pull with a force of 31 N, the sled travels a distance of 53 m. How much work do you do?
Physics
1 answer:
kati45 [8]3 years ago
7 0

Answer:

Explanation:

mass of sled, m = 22 kg

Force of pull, F = 31 N

distance move, d = 53 m

Angle between force and distance, θ = 40°

The formula for the work done is

W = F x d x Cos θ

W = 31 x 53 x cos 40

W = 1258.6 J

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Answer:

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Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
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Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

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3 years ago
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Answer:

28,400 N

Explanation:

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p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa

On the lower part of the hatch, there is a pressure equal to

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F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N

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