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Allisa [31]
3 years ago
8

All celestial objects we tracted to each other by gravity.Asteroids A and B are 5 meters (m) part. Asteroids C and D are also 5

m apart. Asteroids A and B both have a mass of 5 kilograms(kg) and asteroids C and D both have a mass of 7 kg.Between which set of asteroids is the gravitational attraction?
A between asteroids c and d because gravitational attraction is greatest with increased mass

B between asteroids a and b because gravitational attraction is greatest with decreased mass

C between asteroids a and b because gravitational attraction is greatest with increased distance

D between asteroids c and d because gravitational attraction is greatest with increased distance
Physics
1 answer:
Nadusha1986 [10]3 years ago
6 0
A. Between asteroids C and D because gravitational attraction is greatest with increased mass.
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cestrela7 [59]

Answer:

Energy= 46.08KJ

Explanation:

Given that the power needed to light each bulb is 32W

We know that Power = \frac{energy}{time}

The energy needed to light one bulb=power*time

Given time = 1minute = 60 seconds

Energy = 32W*60sec=1920J

Therefore energy needed to light one bulb is 1920J

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8 0
2 years ago
A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle.
chubhunter [2.5K]

Answer:

 v_{1fy} = - 0.4549 m / s

Explanation:

This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved

initial. Before the crash

      p₀ = m v₁₀

final. After the crash

      p_{f} = m v_{1f} + m v_{2f}

Recall that velocities are a vector so it has x and y components

       p₀ = p_{f}

we write this equation for each axis

X axis

       m v₁₀ = m v_{1fx} + m v_{2fx}

       

Y Axis  

       0 = -m v_{1fy} + m v_{2fy}

the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components

      sin 23.3 = v_{2fy} / v_{2f}

      cos 23.3 = v_{2fx} / v_{2f}

      v_{2fy} = v_{2f} sin 23.3

      v_{2fx} = v_{2f} cos 23.3

we substitute in the momentum conservation equation

       m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3

       0 = - m v_{1f} sin θ + m v_{2f} sin 23.3

      1.83 = v_{1f} cos θ + 1.15 cos 23.3

       0 = - v_{1f} sin θ + 1.15 sin 23.3

      1.83 = v_{1f} cos θ + 1.0562

        0 = - v_{1f} sin θ + 0.4549

     v_{1f} sin θ = 0.4549

     v_{1f}  cos θ = -0.7738

we divide these two equations

      tan θ = - 0.5878

      θ = tan-1 (-0.5878)

       θ = -30.45º

we substitute in one of the two and find the final velocity of the incident ball

        v_{1f} cos (-30.45) = - 0.7738

        v_{1f} = -0.7738 / cos 30.45

        v_{1f} = -0.8976 m / s

the component and this speed is

       v_{1fy} = v1f sin θ

       v_{1fy} = 0.8976 sin (30.45)

       v_{1fy} = - 0.4549 m / s

8 0
3 years ago
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