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mestny [16]
3 years ago
5

The international space station travels at a distance of about 250 miles above Earth’s surface and at a speed of 17,500 miles pe

r hour. Even though space dust and debris can be tiny, 1 centimeter or less in size with a mass of 10 grams or less, there is always fear that the space station will sustain damage from a collision with a small piece of space debris.
Why should scientists be concerned about a collision with such a tiny object? Use evidence and scientific reasoning to explain your answer.
Physics
1 answer:
Arisa [49]3 years ago
3 0

Answer:

In this case we are dealing with the pythagorean theorm involving right angled triangles. This theorm states that a^2 + b^2 = c^2 which means the square of the hypotenuse (side c, opposite the right angle) is equal to the square of the remaining two sides.

In this case we will say that a = 3963 miles which is the radius of the earth. c is equal to the radius of the earth plus the additional altitude of the space station which is 250 miles; therefore, c = 4213 miles. We must now solve for the value b which is equal to how far an astronaut can see to the horizon.

(3963)^2 + b^2 = (4213)^2

b^2 = 2,044,000

b = 1430 miles.

The astronaut can see 1430 miles to the horizon.

Explanation:

:D hopes this Helps

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Evgesh-ka [11]

Answer;

D. where two plates collide

Explanation;

-Subduction zones are plate tectonic boundaries where two plates converge, and one plate is thrust beneath the other. This process results in geohazards, such as earthquakes and volcanoes.

-Subduction zone volcanism occurs where two plates are converging on one another. One plate containing oceanic lithosphere descends beneath the adjacent plate, thus consuming the oceanic lithosphere into the earth's mantle. This on-going process is called subduction.

3 0
3 years ago
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Una carga de -10Mc está situada a 20cm delante de otra carga de 5 Mc. Calcular la fuerza electrostática en Newton ejercida por u
tino4ka555 [31]

Answer:

a)  force between them is attraction,   b)  F = 1.125 10⁻² N

Explanation:

In this case the electric force is given by Coulomb's law

          F =k \frac{q_1q_2}{r^2}

           

In the exercise they give us the values ​​of the loads

          q1 = - 10 mC = -10 10⁻³ C

          q2 = 5 mC = 5 10⁻³ C

           d = 20 cm = 0.20 m

let's calculate

          F = 9 10⁹ \frac{10 \ 10^{-3} \ 5 \ 10^{-3}}{0.20^2}

          F = 1.125 10⁻² N

To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction

7 0
3 years ago
A negative charge of -2.0 C and a positive charge of 3.0 C are separated by 80 m. What is the electrostatic force between the tw
faltersainse [42]

Answer:

1. 8437500 N

2. The force between the two charges is attractive.

Explanation:

1. Determination of the force between the two charges.

Charge 1 (q₁) = –2.0 C

Charge 2 (q₂) = 3.0 C

Distance apart (r) = 80 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Force (F) =?

F = Kq₁q₂ / r²

F = 9×10⁹ × 2 × 3 / 80²

F = 5.4×10¹⁰ / 6400

F = 8437500 N

Thus, the force of attraction between the two charges is 8437500 N

2. From the question given, the charges are:

Charge 1 (q₁) = –2.0 C

Charge 2 (q₂) = 3.0 C

We understood that like charges repels while unlike charges attract. Since the two charges (i.e –2 C and 3 C) has opposite signs, it means they will attract each other.

Thus the force between them is attractive.

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