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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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3 years ago
if you exert 10.0 N of force to compress a spring 50 cm from its equilibrium point, then what is the spring constant?
Olegator [25]
The answer would be: b. The spring constant is 20 N/m
4 0
4 years ago
The amount of energy required to raise one mole of a substance one degree Celsius is known as
Yuliya22 [10]

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Heat Capcity

Explanation:

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7 0
3 years ago
An ear squeeze occurs when: Select one: The pressure inside the middle ear space is greater than ambient (surrounding) pressure.
matrenka [14]

B. An ear squeeze occurs when the pressure inside the middle ear space is less than ambient (surrounding) pressure.

<h3>When external ear squeeze occurs</h3>

External ear squeeze occurs when gas trapped in the external ear canal remains at atmospheric pressure while the external water pressure increases during descent.

Thus, we can conclude that, an ear squeeze occurs when the pressure inside the middle ear space is less than ambient (surrounding) pressure.

Learn more about ear squeeze here: brainly.com/question/11430998

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6 0
2 years ago
A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads sh
Dmitry_Shevchenko [17]

Answer:

Explanation:

Let the charge on bead A be q nC  and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = \frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

3 0
3 years ago
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