Answer:
(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.
(II). The torque is 84.87 N-m.
Explanation:
Given that,
Initial spinning = 50.0 rad/s
Time = 20.0
Distance = 2.5 m
Mass of pole = 4 kg
Angle = 60°
We need to calculate the angular acceleration
Using formula of angular velocity
The angular acceleration is -2.5 rad/s²
We need to calculate the number of revolution
Using angular equation of motion
Put the value into the formula
The number of revolution is 500 rad.
(II). We need to calculate the torque
Using formula of torque
Put the value into the formula
Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.
(II). The torque is 84.87 N-m.
I think it's 'C' but I won't know for sure until you let me see the diagram.
Answer:
For elliptical orbits: seldom
For circular orbits: always
Explanation:
We start by analzying a circular orbit.
For an object moving in circular orbit, the direction of the acceleration (centripetal acceleration) is always perpendicular to the direction of motion of the object.
Since acceleration has the same direction of the force (according to Newton's second law of motion), this means that the direction of the force (the centripetal force) is always perpendicular to the velocity of the object.
So for a circular orbit,
the direction of the velocity of the satellite is always perpendicular to the net force acting upon the satellite.
Now we analyze an elliptical orbit.
An elliptical orbit correponds to a circular orbit "stretched". This means that there are only 4 points along the orbit in which the acceleration (and therefore, the net force) is perpendicular to the direction of motion (and so, to the velocity) of the satellite. These points are the 4 points corresponding to the intersections between the axes of the ellipse and the orbit itself.
Therefore, for an elliptical orbit,
the direction of the velocity of the satellite is seldom perpendicular to the net force acting upon the satellite.
The total distance covered by the sound wave is twice the distance between the camera and the subject (because the wave has to reach the subject and then travel back to the camera), so 2L, where L=3.42 m. The speed of sound is v=343 m/s. It is a uniform linear motion, so we can use the basic relationship between space (S), time (t) and velocity (v) to find the time the wave needs to return to the camera:
