Answer: C
high; large
Explanation:
The wave energy is related to its amplitude and frequency.
The wave energy is proportional to the amplitude of the wave. So, wave with the most energy will have high amplitude.
Also, frequency is related to wave energy. The larger the frequency, the more the energy of the wave.
Therefore, The waves with the MOST energy have high amplitudes and large
frequencies.
<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>
A table can be made from minerals but other kinds of solids as well
<span>First, she should put the sample in a test tube and place it in a centrifuge. This would cause the red blood cells to move to the bottom because of their higher density. Next, she would be able to decant the plasma and analyze it separately from the red blood cells.</span>
Answer:
y = 138.96 m
Explanation:
The angle subtended by the moon is the mean of the angle of the arc between the two most extreme points of the moon, we can see that the angle is very small, so we can approximate this arc to a straight line and then use the trigonometric relationships
sin θ = y / L
where L = 15.9 10³ m and θ = 8.74 10⁻³ rad
y = L sin θ
y = 15.9 10³ sin (8.74 10⁻³)
y = 15.9 10³ 0.0087399
y = 138.96 m
