The discoverer of radioactivity was
2 answers:
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Answer:
find the diagram in the attachment.
Explanation:
Let vi = 12 m/s be the intial velocy when the ball is thrown, Δy be the displacement of the ball to a point where it starts returning down, g = 9.8 m/s^2 be the balls acceleration due to gravity.
considering the motion when the ball thrown straight up, we know that the ball will come to a stop and return downwards, so:
(vf)^2 = (vi)^2 + 2×g×Δy
vf = 0 m/s, at the highest point in the upward motion, then:
0 = (vi)^2 + 2×g×Δy
-(vi)^2 = 2×g×Δy
Δy = [-(vi)^2]/2×g
Δy = [-(-12)^2]/(2×9.8)
Δy = - 7.35 m
then from the highest point in the straight up motion, the ball will go back down and attain the speed of 12 m/s at the same level as it was first thrown
Answer:
Explanation:
We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.
Well, according to the law of universal gravitation:
(1)
Where:
is the module of the force exerted between both bodies
is the gravitational constant
is the mass of the Earth
are the mass of each communications satellite
is the distance between the center of the Earth and the satellite
is the radius of the Earth
is the height of the satellite, measured from the Earth's surface
On the other hand, we know according to <u>Newton's 2nd law of motion:</u>
(2)
Combining (1) and (2):
(3)
Isolating :
(4)
Remembering :
(5)
Finally: