(a) Determine the circumference of the Earth through the equation,
C = 2πr
Substituting the known values,
C = 2π(1.50 x 10¹¹ m)
C = 9.424 x 10¹¹ m
Then, divide the answer by time which is given to a year which is equal to 31536000 s.
orbital speed = (9.424 x 10¹¹ m)/31536000 s
orbital speed = 29883.307 m/s
Hence, the orbital speed of the Earth is ~29883.307 m/s.
(b) The mass of the sun is ~1.9891 x 10³⁰ kg.
Answer:
if I am correct, they indicate less steep terrain. think of it as the steeper the terrain the closer together the lines would be. hope that makes sense for you guys.
Answer:
Explanation:
We need only to apply the definition of acceleration, which is:
In our case the final velocity is 
, the initial velocity is 
since it departs from rest, the final time is 
and the initial time we are considering is 
So for our values we have:
Good. You can do some very interesting experiments with that equipment.
Answer:
2.73414 seconds
467622.66798 J
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s² = a
or
The time taken is 2.73414 seconds
The potential energy is given by
The change in potential energy is 467622.66798 J
