F=mg=Gm1m2/r^2
g=Gm2/r^2
g=2Gm2/(2r)^2=2Gm2/4r^2=Gm2/2r^2
So since there is half times the gravity on this unknown planet that has twice earth's mass and twice it's radius, then the person can jump twice as high. 1.5*2= 3m high
Answer:
The correct option is D
Explanation:
In trying to achieve what the student wanted to see, which is to see the relationship between the weight the cord can hold and how long the cord will stretch. Since the origin of the graph is from zero, the value plotted on the vertical axis would be just the length caused by each weights. Thus, <u>the original length would have to be subtracted from the measured length to determine the actual length caused by the weight added to the cord</u>.
Answer:
2.0 m/s/s
Explanation:
The acceleration of an object is the rate of change of velocity of the object.
Mathematically, it is given by:
where
u is the initial velocity
v is the final velocity
t is the time taken for the velocity to change from u to v
Acceleration is a vector, so it has both a magnitude and a direction.
For the runner in this problem, we have:
u = 0 is the initial velocity (he starts from rest)
v = 8.0 m/s is the final velocity
t = 4.0 s is the time taken
Substituting, we find
Answer:
20.96 h
Explanation:
The perimeter of the track is 2*pi*r = 20pi miles
In 10 hours, car B would have moved 20miles. So, when Car A leaves from point X, car B is 20pi - 20 miles from point X counter-clockwise and car A.
From here, we can express the distance of A from X like this:
xa = 3t
And the distance of B would be:
xb = 20pi - 20 - 2t
The time t where they would passed each other and put 12 miles between them would be the one where xa - xb is equal to 12:
xa - xb = 12
3t - (20pi - 20 - 2t) = 12
5t = 20 pi - 8
t = (20pi - 8)/5 = 10.96 h
Remember to add this value to the 10 hours car B had already been racing:
t = 20.96h
