Answer:
The answer is C because there is no friction there will be no friction force only applied and since its on ice you have to account for gravity
Explanation:
The velocity vector of the planet points toward the center of the circle is the following is true about a planet orbiting a star in uniform circular motion.
A. The velocity vector of the planet points toward the center of the circle.
<u>Explanation:</u>
Motion of the planet around the star is mentioned to be uniform and around a circular path. Objects in uniform circular motion motion has constant angular speed but the velocity of the object will not remain constant. Since the planet is in circular motion the direction of velocity vector at a particular point is tangential to the circular path at that particular point.
Thus at every point, the direction of velocity vector changes and this means the velocity is never constant. The objects in uniform circular motion has centripetal acceleration which means that velocity vector of the planet points toward the center of the circle.
The correct answer is A) Ipsilateral
Explanation:
The term ipsilateral is commonly used to describe objects or structures that are on the same side of a body or structure. This term is correct to describe the right eye and the right lung because these two organs are on the same side of the body (the right side). This can also be used to describe other organs such as the left humerus and the left hand or the right ear and the right feet because these pairs are also on the same side. According to this, the correct answer is A.
Answer:
velocity and displacement answer
Explanation:
thanks me
Explanation:
There are three forces on the bicycle:
Reaction force Rp pushing up at P,
Reaction force Rq pushing up at Q,
Weight force mg pulling down at O.
There are four equations you can write: sum of the forces in the y direction, sum of the moments at P, sum of the moments at Q, and sum of the moments at O.
Sum of the forces in the y direction:
Rp + Rq − (15)(9.8) = 0
Rp + Rq − 147 = 0
Sum of the moments at P:
(15)(9.8)(0.30) − Rq(1) = 0
44.1 − Rq = 0
Sum of the moments at Q:
Rp(1) − (15)(9.8)(0.70) = 0
Rp − 102.9 = 0
Sum of the moments at O:
Rp(0.30) − Rq(0.70) = 0
0.3 Rp − 0.7 Rq = 0
Any combination of these equations will work.
