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Inga [223]
1 year ago
7

The gravitational force, Fbetween an object and the Earth is inversely proportional to the square of the distance from the objec

t to the center of the Earth. If an astronaut weighs 214 pounds on the surface of the Earth, what will this astronaut weigh 500 miles above the Earth? Assume that the radius of the Earth is 4000m / l * es Round off your answer to the nearest pound)

Physics
1 answer:
yulyashka [42]1 year ago
7 0

ANSWER:

169 pounds

STEP-BY-STEP EXPLANATION:

We can calculate the weight of the astronaut since we know the relationship between both variables (weight and radius), so we establish the following proportion:

\frac{W_2}{W_1}=\frac{R^2}{(R+d)^2}

The astronaut's initial weight is 214 pounds (W1), the radius is 4000 miles (R) and the distance is equal to 500 miles, we substitute and calculate the new weight of the astronaut as follows:

\begin{gathered} W_2=W_1\cdot\frac{R^{2}}{(R+d)^{2}} \\  \\ \text{ We replacing} \\  \\ W_2=214\cdot\frac{4000^2}{\left(4000+500\right)^2} \\  \\ W_2=169.086=169\text{ pounds} \end{gathered}

The astronaut's new weight is 169 pounds

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3 years ago
A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
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