Answer:
∆L=aL∆T
Explanation:
that's the answer for your Question
Answer:
D
Explanation:
We know the formula for Work to be:
W = f * d
Where W is work done
f is force
d is the distance
A)
Work = 50
Distance = 50
So, Force is:
Force = 50/50 = 1
B)
Work = 400
Distance = 80
Force = 400/80 = 5
C)
Work = 365
Distance = 73
Force = 365/73 = 5
D)
Work = 144
Distance = 16
Force = 144/16 = 9
Hence, D is the situation in which the force applied is the greatest.
P = Power
W = Work
Delta T = Change of Time
So:
P = 35 Joules
Delta T = 70 secs
W = ?
* Joule/Sec = Watt
Answer: The power used to do this task was 0.5 Watts.
The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of
<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N
Then the work done by this force on the crate as it slides down the ramp is
<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J
The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so
<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2
Solve for <em>v</em> :
<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s
