Answer:
y₀ = 1020.3 m
Explanation:
This is a projectile launching exercise, in this case as the package is released its initial vertical velocity is zero.
y = y₀ + 
t - ½ g t²
when it reaches the ground its height is zero
0 = y₀ + 0 - ½ g t²
y₀ = ½ g t²
let's calculate
y₀ = ½ 9.8 14.43²
y₀ = 1020.3 m
Answer:
F = GMmx/[√(a² + x²)]³
Explanation:
The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is
dF = GmdM/L²
Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.
So, the horizontal components add from two symmetrically opposite mass elements dM,
Thus, the horizontal component of the force is
dF' = dFcosФ where Ф is the angle between L and the x axis
dF' = GmdMcosФ/L²
L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.
L = √(a² + x²)
cosФ = x/L
dF' = GmdMcosФ/L²
dF' = GmdMx/L³
dF' = GmdMx/[√(a² + x²)]³
Integrating both sides we have
∫dF' = ∫GmdMx/[√(a² + x²)]³
∫dF' = Gm∫dMx/[√(a² + x²)]³ ∫dM = M
F = GmMx/[√(a² + x²)]³
F = GMmx/[√(a² + x²)]³
So, the force due to the sphere of mass m is
F = GMmx/[√(a² + x²)]³
Answer:
(a) ΔU=747J
(b) γ=1.3
Explanation:
For (a) change in internal energy
According to first law of thermodynamics the change in internal energy is given as
ΔU=Q-W
Substitute the given values
ΔU=970J-223J
ΔU=747J
For(b) γ for the gas.
We can calculate γ by ratio of heat capacities of the gas
γ=Cp/Cv
Where Cp is the molar heat capacity at constant pressure
Cv is the molar heat capacity at constant volume
To calculate γ we first need to find Cp and Cv
So
For Cp
As we know
Q=nCpΔT
Cp=(Q/nΔT)
From relation of Cv and Cp we know that
Cp=Cv+R
Where R is gas constant equals to 8.314J/mol.K
So
So
γ=Cp/Cv
γ=[(37J/mol.K) / (28.687J/mol.K)]
γ=1.3
1. Ask a question
2. Form a hypothesis
3. Experiment
4.Record data
5.Draw Conclusion
6. Share Results
Answer: I am pretty sure that you should pick radio waves.
Explanation: The scientist should use radio waves. I think this because you can use the radio waves to analyze the signals from outer space. This will work much better than anything there, to analyze it the best possible.
The best I could do.
