46 POINTS will mark brainliest if correct
The radius of Mercury's orbit is r = 5.79 x 1010 m and its orbital period is T=88 days. What is the
magnitude of the orbital velocity for the planet around the sun, assuming a circular orbit?
A.7.21 x 103 m/s
B.8.45* 104 m/s
C.4.79 x 104 m/s
D.5.32 x 104 m/s
Answer:
Consider the velocity-time graph attached below.
The velocity-time graph represents the acceleration of a body under a force.
We can see that is the graph that if a child release the ball above the ground at A, it hits the ground at B. Bounces back with a reaches the top again at C, and hits the ground again at D.
The slope of velocity time graph represents acceleration. From A to B, velocity in increasing constantly with respect to time, which means constant acceleration from A to B. AS velocity increase, momentum of the ball also increases, which results in the increase of Kinetic energy.
At B, the ball hits the ground, the velocity decreases, momentum decrease s, because kinetic energy is transferred from the ball to the ground, due to which the ball would not attain the same height after the bounce.
Then the velocity remains negative at C, which means that now the ball is moving in opposite direction till C. It reaches its new at height at C, which is not the same as that of A because of lost in Kinetic Energy, and fall again.
I would say 2 but I don’t want to get you wrong
Answer:
3N/C
Explanation:
Thinking process:
Let, electric field of 4N/C is in the positive x direction, that is + 4 NC
And the the resulting electric field on the x axis at x =2m
Therefore, based on the data above, the net charge must be in the positive direction. Thus, the magnitude is given by the equation:

Now considering that at x = + 4 NC, the net charge will be:

I think its c particles of the wave medium move parallel to the direction of propagation of the wave