Answer:
Explanation:
All objects <u>radiate</u> energy.
Answer:
A. It must be zero
Explanation:
A spacecraft leaves the solar system at a velocity of 1,500 m/s. The net force on this spacecraft is zero. What can we say about the spacecraft's acceleration?
According to Newton's second law
Force = Mass × acceleration
If the net force is zero
0 = mass × acceleration
0 = ma
a = 0/m
a = 0m/s²
this shows that the acceleration will be zero If the net force is zero
<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>
The distance covered by the object between t =4 s and t = 6 s is 4 m
Explanation:
In a velocity-time graph, the distance covered by the object represented can be found by calculating the area under the curve.
Therefore, the distance covered by the object between t = 4 s and t = 6 s is the area under the curve between 4 s and 6 s.
We see that we have to calculate the area of a triangle, with:
Base:
And height:
Therefore, the area is
So, the distance covered by the object is 4 m.
Learn more about distance:
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