Which would have the longer orbital period: a moon 1 million km from the center of Jupiter, or a moon 1 million km from the cent
1 answer:
Answer:
earth
Explanation:
The formula for the orbital period of the moon is given by
As the time period is inversely proportional to the square root of the acceleration due to gravity of the planet.
As the value of acceleration due to gravity on Jupiter is more than the earth, so the period of moon around the earth is large as compared to the period of the moon around the Jupiter when the distance is same.
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11)
12)
13) 50.0 m/s
14) 41.6 m/s
Explanation:
11)
The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by
where
m is the mass of the object
g is the acceleration due to gravity
h is the height relative to the ground
Here in this problem, when the train is at the top, we have:
m = 8325 kg (mass of the train + riders)
(acceleration due to gravity)
h = 127 m (height of the train at the top)
Substituting,
12)
According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:
where
is the kinetic energy at the top
is the potential energy at the top
is the kinetic energy at the bottom
is the potential energy at the bottom
The kinetic energy is the energy due to motion; since the train is at rest at the top, we have
Also, at the bottom the height is zero, so the potential energy is zero
Therefore, we find:
13)
The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by
where
m is the mass of the object
v is the speed of the object
From question 12), we know that the kinetic energy of the train at the bottom is
We also know that the mass is
m = 8325 kg
Therefore, we can calculate the speed of the train at the bottom:
14)
At the top of the second hill, the total mechanical energy of the train is still conserved.
Therefore, we can write again:
where
is the kinetic energy at the top of the 1st hill
is the potential energy at the top of the 1st hill
is the kinetic energy at the top of the 2nd hill
is the potential energy at the top of the 2nd hill
From the previous questions, we know that
and
The height of the second hill is
h = 39 m
So we can also find the potential energy at the second hill:
So, the kinetic energy at the second hill is
And so, the speed is
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows: