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algol [13]
3 years ago
11

An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of 3.14×1

013 N from the Sun. The mass of the Sun is 1.99×1030 kg. How far is the asteroid from the Sun?
Physics
1 answer:
Ksivusya [100]3 years ago
3 0
<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

We have gravitational force

                 F=\frac{GMm}{r^2}

           Where G =  6.67 x 10⁻¹¹ N m²/kg²

                       M = Mass of body 1

                       M = Mass of body 2

                       r = Distance between them

Here we have

                 M = Mass of Sun = 1.99×10³⁰ kg

                 m = Mass of asteroid = 4.00×10¹⁶ kg

                 F = 3.14×10¹³ N

Substituting

                   F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m

The asteroid is 4.11 x 10¹¹ m far from Sun

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A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.
EleoNora [17]

(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

(b) The work done  in firing the projectile is 2,500 J.

<h3>Kinetic energy of the projectile at maximum height</h3>

The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;

K.E = ¹/₂mv₀ₓ²

where;

  • m is mass of the projectile
  • v₀ₓ is the initial horizontal component of the velocity at maximum height

<u>Note:</u> At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.

K.E = (0.5)(2)(30²)

K.E = 900 J

<h3>Work done in firing the projectile</h3>

Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.

W = K.E(i) = ¹/₂mv²

where;

  • v is the resultant velocity

v = √(30² + 40²)

v = 50 m/s

W = (0.5)(2)(50²)

W = 2,500 J

Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

The work done  in firing the projectile is 2,500 J.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

5 0
2 years ago
PLEASE HELP ME ON THIS ONE ANYONE
andreev551 [17]

Answer:

A & B

Explanation:

A & B Would be the right answer since Morse code cannot be represented through the height of the fire.

6 0
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What is the net force needed to accelerate a 21.4 kg grocery sack upward at 1.5 m/s2?
marshall27 [118]

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snow_lady [41]

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1 watt means 1 joule per second

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