Answer:
Gravity and
Air resistance
Explanation:
The two forces acting on a skydiver are gravitational force and air resistance.
Gravitational force is a force that tends to pull all massive bodies towards the center of the earth. It works on all bodies that has mass. The larger or bigger the mass, the more the pull of gravity on the body.
Air resistance is the drag of air on a body as it passes to it. It is resisting force.
- When a sky diver jumps out of a plane, he/she encounters both gravity and air resistance.
- It soon balances both force and attain terminal velocity.
- Air resistance is a frictional force that opposes motion.
- This frictional force pushes in the opposite direction of motion
- Motion direction is downward due to the celerity caused by gravity.
Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
Assuming Earth's gravity, the formula for the flight of the particle is:
<span>s(t) = -16t^2 + vt + s = -16t^2 + 144t + 160. </span>
<span>This has a maximum when t = -b/(2a) = -144/[2(-16)] = -144/(-32) = 9/2. </span>
<span>Therefore, the maximum height is s(9/2) = -16(9/2)^2 + 144(9/2) + 160 = 484 feet. </span>
Answer:
0
Explanation:
the momentum will always be 0 when it is at rest because the object isnt moving!
Hope this helped!
Answer:
<h2>3.36J</h2>
Explanation:
Step one:
given data
mass m= 1.3kg
distance moved s= 2.8m
opposing frictional force= 0.34N
assume g= 9.81m/s^2
we know that work done= force *distance moved
1. work done to push the book= 1.55*2.8=4.34J
2. Work against friction = force of friction x distance
= 0.34*2.8=0.952J
Step two:
the work done on the book is the net work, which is
Network done= work done to push the book- Work against friction
Network done= 4.32-0.952=3.36J
<u>Therefore the work of the 1.55N 3.36J</u>
