Correct Answers is A.
The machines gives us some mechanical advantage. This means the mechanical average makes the work output greater than the work input
Simple most example is a lever. The force applied is smaller and the output work is larger as compared to input.
Option B cannot be true, as there must be a force to get some work done.
Option C and D are inverse of what a machine is designed for. A small force can be exerted through a large distance to have a large force exerted through a small distance. Common Example of this principle is a screw opener.
Answer:
At 400 m the potential energy of the mountain climber doubled the initial value.
Explanation:
Given;
initial height of the mountain climber = 200 m
final height of the mountain climber, = 400 m
The potential energy of the mountain climber is calculated as;
Potential energy, P.E = mgh
At 200 m, P.E₁ = mg x 200 = 200mg
At 400 m, P.E₂ = mg x 400 = 400mg
Then, at 400 m, P.E₂ = 2 x 200mg = 2 x P.E₁
Therefore, at 400 m the potential energy of the mountain climber doubled the initial value.
In order to answer this exercise you need to use the formulas
S = Vo*t + (1/2)*a*t^2
Vf = Vo + at
The data will be given as
Vf = final velocity = ?
Vo = initial velocity = 1.4 m/s
a = acceleration = 0.20 m/s^2
s = displacement = 100m
And now you do the following:
100 = 1.4t + (1/2)*0.2*t^2
t = 25.388s
and
Vf = 1.4 + 0.2(25.388)
Vf = 6.5 m/s
So the answer you are looking for is 6.5 m/s
a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.
