A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled
1 answer:
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Answer:
D. 10 T
Explanation:
When a particle is moving in a magnetic field, the magnetic force provides the centripetal force that keeps the particle in circular motion.
The cyclotron period (the period the particle takes to complete one orbit) can be found to be
where
m is the mass of the particle
q is its charge
B is the magnetic field
As we see, the period is directly proportional to the mass of the particle.
In this problem, the second particle is ten times as massive as the first one:
m' = 10 m
while the speed is the same. So, the period of the second particle is
Answer:
a = 2.72 [m/s2]
Explanation:
To solve this problem we must use the following kinematics equation:
where:
Vf = final velocity = 1200 [km/h]
Vo = initial velocity = 25 [km/h]
t = time = 2 [min] = 2/60 = 0.0333 [h]
1200 = 25 + (a*0.0333)
a = 35250.35 [km/h2]
if we convert these units to units of meters per second squared
Answer:
a) , b)
Explanation:
The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:
Where is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:
Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:
a) t = 50 s.
b) t = 100 s.
Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:
Since , then the angular velocity is equal to zero. Therefore: