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Deffense [45]
4 years ago
6

The gravitational force, F, between an object and the Earth is inversely proportional to the square of the distance from the obj

ect and the center of the Earth. If an astronaut weighs 220 pounds on the surface of the Earth, what will this astronaut weigh 2250 miles above the Earth?
Physics
1 answer:
PolarNik [594]4 years ago
8 0

Answer:

398.0 N = 89.74 pound force

Explanation:

Gravitational force:

F=G\frac{mM}{d^2}

R = 6371 km (Radius of the Earth)

At the surface, weight of the surface w =220 pounds = 978.61 N

2250 miles = 3621024 m

total distance from the center of the Earth

3621024 m + 6.371×10⁶ m = 9.99×10⁶ m

\frac{w'}{w}=\frac{R^2}{(R+d)^2}\\w'=\frac{(6.371\times10^6)^2}{(9.99\times10^6)^2}\times 978.61N=398.0 N

398.0 N = 89.74 pound force

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A car starts from rest at a stop sign. It accelerates at 4.0 m/s^2 for 3 seconds, coasts for 2 s, and then slows down at a rate
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<span>FIRST SECTION You should use the formula for uniformly accelerated linear movement. Initial speed is 0 because it starts from rest. d=(1/2)*a*t^2+vo*t =(1/2)*(4.0 m/s^2)*(3s)^2+0*3s=(1/2)*(4.0 m/s^2)*3^2*s^2+0=2.0 m*9=18m You can calculate the final speed with the other formula: v=a*t+vo=(4.0 m/s^2)*(3s)+0=(4.0 m/s)*(3)=12m/s SECOND SECTION You should use the formula for uniform linear movement. Velocity is a constant: it remains in 12m/s. d=v*t=12m/s*2s=12m*2=24m THIRD SECTION We should use the same formulas as the first section, but with different numbers. Initial velocity will be 12m/s, and then velocity will start to decrease until it gets to 0. We don’t know what the time is for this section. Acceleration is negative, because it’s slowing down. v=a*t+vo 0=-3.0 m/s^2*t+12m/s 3.0 m/s^2*t=12m/s t=(12m/s)/(3.0 m/s^2)=4(1/s)/(1/s^2)=4s^2/s=4s Now let’s use that time in the other formula: d=(1/2)*a*t^2+vo*t =(1/2)*(-3.0 m/s^2)*(4s)^2+(12m/s)*3s=(-1.5 m/s^2)*4^2*s^2+12*3m*s/s=-1.5 m*4^2+36m=-1.5*16m+36m=-24m+36m=12m Now let’s add the 3 stages: d=18m+24m+12m=54m </span>
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D

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