Refer to the diagram shown below.
u = 0, the initial vertical velocity
Assume g = 9.8 m/s² and ignore air resistance.
At the first stage of landing on the ground, the distance traveled is
h = 3.1 - 0.6 = 2.5 m.
If v = the vertical velocity at this stage, then
v² = u² + 2gh
v² = 2*(9.8 m/s²)*(2.5 m) = 49 (m/s)²
v = 7 m/s
At the second stage of landing on the ground, let a = the acceleration (actually deceleration) that his body provides to come to rest.
The distance traveled is 0.6 m.
Therefore
0 = (7 m/s)² + 2(a m/s²)*(0.6 m)
a = - 49/1.2 = - 40.833 m/s²
Answers:
(a) The velocity when the man first touches the ground is 7.0 m/s.
(b) The acceleration is -40.83 m/s² (deceleration of 40.83 m/s²) to come to rest within 0.6 m.
Answer:
8.33 minutes
Explanation:
First let us find out how much time it takes for the light for the sun to reach the earth
Distance between Earth and Sun is 1.5×10¹¹ m
Speed of light = 3×10⁸ m/s

Converting to minutes

So, it takes 8.33 minutes for the light from the sun to reach Earth.
This means that we would be receiving light for 8.33 minutes after the Sun turned off.
Choose the element that has a smaller atomic radius :scandium
Answer:
(a). Energy is 64,680 J
(b) velocity is 51.43m/s
(c) velocity in mph is 115.0mph
Explanation:
(a).
The potential energy
of the payload of mass
is at a vertical distance
is
.
Therefore, for the payload of mass
at a vertical distance of
, the potential energy is


(b).
When the payload reaches the bottom of the shaft, all of its potential energy is converted into its kinetic energy; therefore,




(c).
The velocity in mph is

