The centripetal force (of gravity) on a satellite in orbit is an
unbalanced force (because there's no equal force pulling
the satellite away from Earth), changes the direction of the
satellite (into a closed orbit instead of a straight line), and
always acts toward the center of whatever curve the satellite
happens to be on at the moment.
Answer:
fixed pulley: A pulley system in which the pulley is attached to a fixed point and the rope is attached to the object. ... movable pulley: A pulley system in which the pulley is attached to the object; one end of the rope is attached to a fixed point and the other end of the rope is free.
Explanation:
Hahahahha ok it’s B or C or it B
Answer:
Explanation:
Work is equal to the product of force and distance.
The force is 8 Newtons and the distance is 15 meters.
Substitute the values into the formula.
Multiply.
- 1 Newton meter is equal to 1 Joule
- Our answer of 120 N*m equals 120 J
The work done is <u>120 Joules</u>
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
