The amusement park ride shown above takes riders straight up a tall tower and then releases an apparatus holding seats. This app
aratus free-falls back to Earth and is stopped safely right above the ground. Which of the following indicates the magnitude of the gravitational force exerted on a rider of mass m on the way up and on the way down?
1 answer:
The gravitational force on the rider is:
- Way up: equal to mg
- Way down: equal to mg
Explanation:
The choices are missing: find them in the attached figure.
The force of gravity acting on a body is a force directed downward (towards the Earth's centre) and whose magnitude is
where
m is the mass of the body
g is the acceleration due to gravity
The value of g is approximately constant near the Earth's surface and it is
During the ride, the mass of the rider, m, remains constant. This means that the magnitude of the gravitational force, mg, exerted on the rider remains constant during the rider.
Therefore, the correct answer is
- Way up: equal to mg
- Way down: equal to mg
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Answer:
719
Explanation:
Conversion
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Answer:
<h2>187,500N/m</h2>Explanation:
From the question, the kinectic energy of the train will be equal to the energy stored in the spring.
Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².
Equating both we will have;
1/2 mv² = 1/2ke²
mv² = ke²
m is the mass of the train
v is the velocity of then train
k is the spring constant
e is the extension caused by the spring.
Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m
Substituting this values into the formula will give;
30000*4² = k*1.6²
The value of the spring constant is 187,500N/m