Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated
1 answer:
Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN
Explanation: To find the answer we need to know more about the Newton's law of gravitation.
<h3>What is Newton's law of gravitation?</h3>- Gravitation is the force of attraction between any two bodies.
- Every body in the universe attracts every other body with a force.
- This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
- Mathematically we can expressed it as,
- Here, we have given with the data's,
- Thus, the force of attraction between these two bodies will be,
Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.
Learn more about the Newton's law of gravitation here:
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The box has 3 forces acting on it:
• its own weight (magnitude <em>w</em>, pointing downward)
• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)
• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)
Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have
• net parallel force:
∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>
• net perpendicular force:
∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0
Solve the net perpendicular force equation for the normal force:
<em>n</em> = <em>w</em> cos(35°)
<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)
<em>n</em> ≈ 120 N
Solve for the mag. of friction:
<em>f</em> = <em>µ</em> <em>n</em>
<em>f</em> = 0.25 (120 N)
<em>f</em> ≈ 30 N
Solve the net parallel force equation for the acceleration:
-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>
<em>a</em> ≈ (54.3157 N) / (15 kg)
<em>a</em> ≈ 3.6 m/s²
Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:
<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>
<em>v</em>² = 2 (3.6 m/s²) (3 m)
<em>v</em> = √(21.7263 m²/s²)
<em>v</em> ≈ 4.7 m/s
