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Nikolay [14]
3 years ago
13

Assume the earth to be a nonrotating sphere with mass MEME and radius RERE. If an astronaut weights WW on the ground, what is hi

s weight when he is 2RE2RE above the surface of the earth
Physics
1 answer:
Solnce55 [7]3 years ago
6 0

Answer:

The weight at a distance 2 RE from surface of earth is <em>W/9</em>

Explanation:

For the value of acceleration due to gravity (g), we have a formula, that is:

g = (G)(ME)/(RE)²    ----- equation (1)

where,

G = Gravitational Constant

ME = Mass of Earth

RE = Radius of Earth

g = Acceleration due to gravity on surface of earth = 9.8 ms²

When the person goes 2RE, distance above earth's surface. Then the total distance from center of earth becomes: 2RE + RE = 3RE.

Therefore, equation (1) becomes:

gh = (G)(ME)/(3RE)²

where,

gh = acceleration due to gravity at height

gh = (G)(ME)/(RE)²9

using equation (1), we get:

gh = g/9

Now, he weight is given by formula:

W = mg   ------- equation (2)

At height 2RE

Wh = (m)(gh)

where,

Wh = Weight at height = ?

m = mass of astronaut

Therefore, using vale of gh, we get:

Wh = mg/9

Using equation (2), we get:

<u>Wh = W/9</u>

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3 years ago
if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu
RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

F = \frac{k|q_1||q_2|}{r^2}

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

Therefore, the mutual force between the two point charges is 319.64 N

4 0
3 years ago
Which of the following statements are true? Positively charged objects attract other positively charged objects. Negatively char
butalik [34]
The statement that is true is that positively charged objects attract negatively charged objects. This is due to a law that states 'like forces attract while unlike forces repel. This same concept applies to magnetism. If you put two similar poles together, for example; if you place two south poles together. You feel a separating force between the two poles. But if you place two opposite poles together they attract each other. Hope i helped. <span />
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3 years ago
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Please help ASAP!!
inessss [21]

Answer:

at t=46/22, x=24 699/1210 ≈ 24.56m

Explanation:

The general equation for location is:

x(t) = x₀ + v₀·t + 1/2 a·t²

Where:

x(t) is the location at time t. Let's say this is the height above the base of the cliff.

x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0

v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.

a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².

Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.

Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²

Stone: x(t) = 0 + 22·t - 1/2*9.8 t²

Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:

46 = 22·t

so t = 46/22 ≈ 2.09

Put this t back into either original (i.e., with the quadratic term) equation and get:

x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m

4 0
4 years ago
As the train in the image moves to the right how does the train horn sound to person a?
alina1380 [7]

Answer:

Explanation:

Person A's velocity relative to the train is 0.  Therefore, the pitch of the horn will not change.

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