Answer:
D. Graphing the force as a function of distance and calculating the area under the curve.
Explanation:
<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>
Answer:
The change in velocity is 15.83 [m/s]
Explanation:
Using the Newton's second law we have:
ΣF = m*a
The force in the graph is 185 N, therefore:
![185=0.369*a\\Where\\a=acceleration made it by the force [m/s^2]](https://tex.z-dn.net/?f=185%3D0.369%2Aa%5C%5CWhere%5C%5Ca%3Dacceleration%20made%20it%20by%20the%20force%20%5Bm%2Fs%5E2%5D)
![a=501.35[m/s^2]](https://tex.z-dn.net/?f=a%3D501.35%5Bm%2Fs%5E2%5D)
Now using the following kinematic equation:
![V^{2}=Vi^{2} + 2*a*(x-xi) \\where\\V=final velocity [m/s]\\Vi= initial velocity [m/s] = 0 the hockey disk is in rest when receives the hit.\\ x = Final position [m] = 0.4 m\\xi = initial position [m] = 0.15m\\](https://tex.z-dn.net/?f=V%5E%7B2%7D%3DVi%5E%7B2%7D%20%2B%202%2Aa%2A%28x-xi%29%20%5C%5Cwhere%5C%5CV%3Dfinal%20velocity%20%5Bm%2Fs%5D%5C%5CVi%3D%20initial%20velocity%20%5Bm%2Fs%5D%20%3D%200%20the%20hockey%20disk%20is%20in%20rest%20when%20receives%20the%20hit.%5C%5C%20x%20%3D%20Final%20position%20%5Bm%5D%20%3D%200.4%20m%5C%5Cxi%20%3D%20initial%20position%20%5Bm%5D%20%3D%200.15m%5C%5C)
Now replacing the values:
![V^{2}=0 + 2*501.35*(0.4-0.15)\\ \\V= 15.83[m/s]](https://tex.z-dn.net/?f=V%5E%7B2%7D%3D0%20%2B%202%2A501.35%2A%280.4-0.15%29%5C%5C%20%5C%5CV%3D%2015.83%5Bm%2Fs%5D)
F = 2820.1 N
Explanation:
Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as
Fnet = ma = 0 (a = 0 no sliding)
= F - mgsin15°
= 0
or
F = mgsin15°
= (120 kg)(9.8 m/s^2)sin15°
= 2820.1 N
Answer: Yes
because.....
When the cruise control is engaged, the throttle can still be used to accelerate the car. Also,
* Hopefully this helps:) Mark me the brainliest:)!!!
