An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid
1 answer:
Explanation:
It is given that, An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. We need to find the ratio of their masses.
As they are in equilibrium, the force of gravity due to each other is same. So,
So, the ratio of masses X/Y is 0.0847
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Answer:
1. The final velocity of the penny before it hits the ground is approximately 54.25 m/s
2. The velocity after falling 45 meters is approximately 37.10 m/s
3. The height up the hill one can start without going over the smaller hill is approximately 2.75 meters
Explanation:
The height of the Bellagio, h = 150 meters
The mass of the penny, m = 1 kg
The kinematic equation of motion that can be used to find the final velocity of the penny 'v' before it hits the ground, is presented as follows;
v² = u² + 2·g·h
Where;
v = The final velocity of the penny after dropping through a height, 'h'
u = The initial velocity of the penny = 0 m/s for the penny initially at rest
g = The acceleration due to gravity ≈ 9.81 m/s²
h = The height from which the penny was dropped = 150 m
∴ v² ≈ 0² + 2 × 9.81 × 150 = 2,943
v ≈ √2,943 ≈ 54.25
The final velocity of the penny before it hits the ground, v ≈ 54.25 m/s
2. Here, the initial velocity, u = 80 km/h = 80 km/h × 1000 m/km × 1 h/(60 × 60 s) = 200/9 m/s = 22. m/s
The height of supreme scream, = 90 meters
The height at which the velocity is required, h = 45 meters
From v² = u² + 2·g·h, we get;
v² = 22.² + 2 × 9.81 × 45 ≈ 1,376.73
∴ v = √1,376.73 ≈ 37.10
The velocity 'v' after falling 45 meters is, v = 37.10 m/s
3. The height of the smaller hill, h = 5 meters
The running start = 4 m/s = The initial velocity
The velocity required to reach the height, h, of the smaller heal v = √(2·g·h)
∴ v = √(2 × 9.81 m/s² × 5 m) ≈ 9.9 m/s
The height 'h'' up the larger hill that will give a velocity, 'v', at the bottom of the smaller hill of approximately 9.9 m/s with an initial velocity, u = 4 m/s, is given as follows;
v² = u² + 2·g·h'
9.9² = 4² + 2 × 9.81 × h'
∴ h' = 9.9²/(4² + 2 × 9.81) ≈ 2.75
Given that the running start is 40 m/s, the height up the hill one can start without going over the smaller hill, h' ≈ 2.75 meters