Answer:
U
Explanation:
The first law of thermodynamics states that:
where
U is the internal energy of the gas, which represents the sum of the chemical and thermal energy stored in the atoms and molecules of the gas
represents the variation of internal energy
Q is the heat absorbed by the system
W is the work done by the system
So, the sum of the chemical and thermal energy stored in atoms and molecules is represented by U, the internal energy.
Answer / Explanation
It is worthy to note that the question is incomplete. There is a part of the question that gave us the vale of V₀.
So for proper understanding, the two parts of the question will be highlighted.
A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s later. You may ignore air resistance.
a) What must the height of the building be for both balls to reach the ground at the same time if (i) V₀ is 6.0 m/s and (ii) V₀ is 9.5 m/s?
b) If Vo is greater than some value Vmax, a value of h does not exist that allows both balls to hit the ground at the same time.
Solve for Vmax
Step Process
a) Where h = 1/2g [ (1/2g - V₀)² ] / [(g - V₀)²]
Where V₀ = 6m/s,
We have,
h = 4.9 [ ( 4.9 - 6)²] / [( 9.8 - 6)²]
= 0.411 m
Where V₀ = 9.5m/s
We have,
h = 4.9 [ ( 4.9 - 9.5)²] / [( 9.8 - 9.5)²]
= 1152 m
b) From the expression above, we got to realise that h is a function of V₀, therefore, the denominator can not be zero.
Consequentially, as V₀ approaches 9.8m/s, h approaches infinity.
Therefore Vₙ = V₀max = 9.8 m/s
The steps<span> of the </span>scientific method<span> are to: Ask a Question. Do Background Research. Construct a Hypothesis. Test Your Hypothesis by Doing an Experiment. Analyze Your Data and Draw a Conclusion.</span>
The intensity of the sound wave is defined as the ratio between the power of the wave and the area through which the wave passes:
where
I is the intensity
P is the power
A is the area
If we use the data of the problem,
and
, we find the intensity of the sound wave:
==> Object A travels for 60 seconds before Object B starts out.
==> Object A moves at 2 m/s.
==> So Object A has a lead of 120 m when Object B starts out.
==> Object B moves at 3 m/s . . . 1 m/s faster than Object A.
==> So Object B catches up on Object A by 1 m every second.
==> Object B closes up Object A's lead of 120 m in <em>120 seconds</em>.
