<span>118 C
The Clausius-Clapeyron equation is useful in calculating the boiling point of a liquid at various pressures. It is:
Tb = 1/(1/T0 - R ln(P/P0)/Hvap)
where
Tb = Temperature boiling
R = Ideal Gas Constant (8.3144598 J/(K*mol) )
P = Pressure of interest
Hvap = Heat of vaporization of the liquid
T0, P0 = Temperature and pressure at a known point.
The temperatures are absolute temperatures.
We know that water boils at 100C at 14.7 psi. Yes, it's ugly to be mixing metric and imperial units like that. But since we're only interested in relative pressure differences, it's safe enough. So
P0 = 14.7
P = 14.7 + 12.3 = 27
T0 = 100 + 273.15 = 373.15
And for water, the heat of vaporization per mole is 40660 J/mol
Let's substitute the known values and calculate.
Tb = 1/(1/T0 - R ln(P/P0)/Hvap)
Tb = 1/(1/373.15 K - 8.3144598 J/(K*mol) ln(27/14.7)/40660 J/mol)
Tb = 1/(0.002679887 1/K - 8.3144598 1/K ln(1.836734694)/40660)
Tb = 1/(0.002679887 1/K - 8.3144598 1/K 0.607989372/40660)
Tb = 1/(0.002679887 1/K - 5.055103194 1/K /40660)
Tb = 1/(0.002679887 1/K - 0.000124326 1/K)
Tb = 1/(0.002555561 1/K)
Tb = 391.3034763 K
Tb = 391.3034763 K - 273.15
Tb = 118.1534763 C
Rounding to 3 significant figures gives 118 C</span>
<span>Matching the boundary with its characteristics
1. Convergent - C. Compression
2. Divergent - B. Along ocean ridges
3. Transform - A. Along strike-slip faults
The compression that occur in the convergent boundary causes the reverse fault in the earth crust.
So in the divergent boundary two crust plates move apart causing a normal fault along the ocean ridges.
The faults in the transform boundary happens at the place where plates slide laterally.</span>
Answer:
The percentage of its mechanical energy does the ball lose with each bounce is 23 %
Explanation:
Given data,
The tennis ball is released from the height, h = 4 m
After the third bounce it reaches height, h' = 183 cm
= 1.83 m
The total mechanical energy of the ball is equal to its maximum P.E
E = mgh
= 4 mg
At height h', the P.E becomes
E' = mgh'
= 1.83 mg
The percentage of change in energy the ball retains to its original energy,
ΔE % = 45 %
The ball retains only the 45% of its original energy after 3 bounces.
Therefore, the energy retains in each bounce is
∛ (0.45) = 0.77
The ball retains only the 77% of its original energy.
The energy lost to the floor is,
E = 100 - 77
= 23 %
Hence, the percentage of its mechanical energy does the ball lose with each bounce is 23 %
Doing a force balance on the car:
ma = Fr
ma = μmg
a = μg
a = 0.3(9.81)
a = 29.43 m/s2
Using the formula:
2ax = v2
2(29.43)(34) = v2
v = 44.74 m/s = 161.05 km/h
The car was going 44.74 m/s or 161.05 kph when the brakes were applied.
Answer:
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Explanation:
44e
