Radius of Earth is given as
now here the height from the surface of earth is same as that of radius
now here the acceleration due to gravity at this height is given as
now the force of gravity on the given object will be
<em>so the force of gravity on it is 1.225 N</em>
I would say light. Hope this helps
Answer:
4.2 m/s
Explanation:
The velocity-time graph is piecewise linear. The acceleration in each of the three segments of the graph is uniform. The instant lies between and t = 6.0s 100 s, so the acceleration must be calculated using the slope of the middle segment.
a =
(9.6 -2.4)m/s
------------------
(10.0 -6.0)s
= 1.8 m/s2
The instantaneous velocity is to be found after the object accelerates over an interval T = (7.0 - 6.0) s = 1.0 s, starting from a velocity of 2.4 m/s,
So the velocity at t = 7.0 s is
v = u + aT = 2.4 m/s + (1.8 m/s2)(1.0 s) = 4.2 m/s
The answer is B. Snake.
Hope this helps!
Answer:1) the total distance is the sum of the two distances
60 km + 45 km = 105 km
2) The displacement is the net movement, or the difference between the initial position and the final position
Call x the initial position, then the final position is x + [60km - 45km]
And the displacement is x + (60km - 45km) - x =60km -45 km = 15 km
Explanation:
