Answer: 5,640 s (94 minutes)
Explanation:
the tangential speed of the HST is given by
(1)
where
is the length of the orbit
r is the radius of the orbit
T is the orbital period
In our problem, we know the tangential speed: 
. The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:
So, we can re-arrange equation (1) to find the orbital period:
Dividing by 60, we get that this time corresponds to 94 minutes.
Answer:
A) s=1/2at^2
t=√(2s/a)=√(2x400)/10.0)=9.0s
B) v=at
v=10.0x9=90m/s
Answer:
Explanation:
We define the linear density of charge as:
Where L is the rod's length, in this case the semicircle's length L = πr
The potential created at the center by an differential element of charge is:
where k is the coulomb's constant
r is the distance from dq to center of the circle
Thus.
Potential at the center of the semicircle
It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.
Like using the first law of motion we can determine the force of gravity acting on the cart that has moved a certain distance and the velocity or the speed of card has already been registered and since time is known putting the values in formula would help us calculate the gravitational pull acting on cart.
Answer:
The answer should be C. slanted upward to the right.
Hope this helps. :-)
