Answer:
False
Explanation:
Think of the electric potential in terms of potential energy. If you imagine a place with high elevation (A) and another one at sea level (B), a ball will roll from high potential to low potential (A-->B).
Everything in our universe wants to reach a lower state of energy if no external force is acted upon it. Every object tends to slow down (friction), a radioactive element dissipates energy (an unstable element releases energy to get to a stable state), water in the clouds comes down to the ground (rain experiencing difference in potential energy).
Electric potential is exactly the same, you just can't see it! It flows from higher voltage (which is a synonym for electric potential) to lower voltage.
Substance A would have a delta T (change in temp) rise 1/2 the rise in substance B.
Q=mc x delta T
Q= heat energy in Joules
m= mass of substance heated or cooled
c= specific heat
ΔT is change in temp.
Solve for change in temp=. Q/mc
Specific heat and mass are not inversely proportional to heat energy input.
Putting into real world scenario of using water to heat a building.
Specific heat of water is 1.
It takes 1 btu to raise one pound of water 1 degF. at a base of 60 degF
Acetone specific heat is .51
So it takes half the amount of heat input to get a 100 degree ΔT, as compared to water.
<span>Since there is no friction, conservation of energy gives change in energy is zero
Change in energy = 0
Change in KE + Change in PE = 0
1/2 x m x (vf^2 - vi^2) + m x g x (hf-hi) = 0
1/2 x (vf^2 - vi^2) + g x (hf-hi) = 0
(vf^2 - vi^2) = 2 x g x (hi - hf)
Since it starts from rest vi = 0
Vf = squareroot of (2 x g x (hi - hf))
For h1, no hf
Vf = squareroot of (2 x g x (hi - hf))
Vf = squareroot of (2 x 9.81 x 30)
Vf = squareroot of 588.6
Vf = 24.26
For h2
Vf = squareroot of (2 x 9.81 x (30 – 12))
Vf = squareroot of (9.81 x 36)
Vf = squareroot of 353.16
Vf = 18.79
For h3
Vf = squareroot of (2 x 9.81 x (30 – 20))
Vf = squareroot of (20 x 9.81)
Vf = 18.79</span>
A force sets an object in motion when the force is multiplied by the time of its application.
Answer:
Explanation:
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