gravitational potential is directly proportional to the height of the object relative to a reference line and is given as
PE = mgh
where m = mass of object , g = acceleration due to
gravity and h = height of the object above the reference line .
as the skydiver falls , its height above the ground decrease and hence the gravitational potential energy of the skydiver decrease.
as per conservation of energy , total energy of the skydiver must remain constant all the time . hence the decrease in potential energy appears as increase in kinetic energy by same amount to keep the total energy constant
KE + PE = Total energy
so as the skydiver falls , it gains speed and hence the kinetic energy of skydiver increase since kinetic energy is directly proportional to the square of the speed.
when the parachute opens, the skydiver experience force in upward which tries to balance the weight of the skydiver. hence the speed of the skydiver decrease until upward force becomes equal to the downward force. hence the kinetic energy decrease just after the parachute opens
Answer:
B - A
Explanation:
For the combination of 2 vector to due southwest, 1 vector must due south and the other vector due west. Since vector B is already due west, vector A should due south. As vector A is already due north, vector -A would due south. So the combination of B + (-A) or B - A should points southwest
Answer:
Work done by the gardner is 500 J
Explanation:
As we know that the gardner apply force perpendicular upward by magnitude 300 N and along the floor horizontal force is 100 N
so we have
now the displacement of the gardner along the floor is
now work done is given as
so we have
<h2>
Horizontal component of the rock’s velocity when it strikes the ground is 17.25 m/s</h2>
Explanation:
In horizontal direction there is no acceleration or deceleration for a rock projected at an initial angle of 37° off the ground.
So the horizontal component of velocity always remains the same.
Horizontal component of velocity is the cosine component of velocity.
Initial velocity, u = 21.6 m/s
Angle, θ = 37°
Horizontal component of velocity = u cosθ
Horizontal component of velocity = 21.6 cos37
Horizontal component of velocity = 17.25 m/s
Since the horizontal velocity is unaffected, we have
Horizontal component of the rock’s velocity when it strikes the ground = 17.25 m/s