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Elena L [17]
3 years ago
12

The acceleration of gravity on Earth is approximately 10 m/s2 (more precisely, 9.8 m/s2). If you drop a rock from a tall buildin

g, about how fast will it be falling after 3 seconds?
Physics
1 answer:
Amiraneli [1.4K]3 years ago
4 0

Given Information:  

Acceleration of gravity = g = 9.8 m/s²

Time period = t = 3 seconds

Required Information:

velocity = v = ?

Answer:  

v = 29.4 m/s

Explanation:  

As we know the velocity of an object under free fall assuming that the object was initially at rest is given by

v = -gt

Where minus sign indicates that the object is moving downward

v = -9.8*3

v = -29.4 m/s

Therefore, the rock will be falling at the speed of 29.4 m/s after 3 seconds.

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An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from 10.0∘C to 25.0∘C at constant pressure. The gas do
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Answer:

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Explanation:

For (a) change in internal energy

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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)

C_{p}=\frac{970J}{1.75mol*(25^{o}C-10^{o}C )}\\C_{p}=37J/mol.K

From relation of Cv and Cp we know that

Cp=Cv+R

Where R is gas constant equals to 8.314J/mol.K

So

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γ=1.3

4 0
3 years ago
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