Answer:

Explanation:
We know that the gravity on the surface of the moon is,
<u>Gravity at a height h above the surface of the moon will be given as:</u>
..........................(1)
where:
G = universal gravitational constant
m = mass of the moon
r = radius of moon
We have:
is the distance between the surface of the earth and the moon.
Now put the respective values in eq. (1)

is the gravity on the moon the earth-surface.
Here is the energy that is left after the quantity of energy is transformed: 750 j of electrical energy is changed into 400 j of kinetic or mechanical energy, which is then turned into 0.32 j of efficient energy.
To run the fan, electrical energy is utilized.
Here, under the specified circumstances, 750 J of electrical energy is utilized to operate the fan, which is transformed into 400 J of kinetic energy. As a result, 350 J of energy is wasted due to various frictional and resistive losses.
Therefore, we may conclude that only 400 J of the 750 J available energy is used to power the fan, with the remaining energy being wasted as a result of friction.
Additionally, we can state that this fan's effectiveness will be
n = Useful ÷ Total
n = 400 ÷ 750
n = 8 ÷ 25
n = 0.32
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Answer:
volume measured by pid^3 over 6 i think
Explanation:
Answer:
A) a = 2.31[m/s^2]; B) t = 14.4 [s]
Explanation:
We can solve this problem using the kinematic equations, but firts we must identify the data:
Vf= final velocity = take off velocity = 120[km/h]
Vi= initial velocity = 0, because the plane starts to move from the rest.
dx= distance to run = 240 [m]
![v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\ a=2.31[m/s^2]\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%20%3Dv_%7Bi%7D%20%5E%7B2%7D%2B2%2Ag%2Adx%5C%5Cwhere%3A%5C%5Cv_%7Bf%7D%3D120%5B%5Cfrac%7Bkm%7D%7Bh%7D%20%5D%2A%5Cfrac%7B1hr%7D%7B3600sg%7D%20%2A%20%5Cfrac%7B1000m%7D%7B1km%7D%20%3D33.33%5Bm%2Fs%5D%5C%5C%5C%5CReplacing%5C%5C33.33%5E%7B2%7D%3D0%2B2%2Aa%2A%28240%29%5C%5C%20a%3D%5Cfrac%7B11108.88%7D%7B2%2A240%7D%5C%5C%20%20a%3D2.31%5Bm%2Fs%5E2%5D%5C%5C)
To find the time we must use another kinematic equation.
![v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bi%7D%20%2Ba%2At%5C%5Creplacing%3A%5C%5C33.33%3D0%2B%282.31%2At%29%5C%5Ct%3D%5Cfrac%7B33.33%7D%7B2.31%7D%5C%5C%20t%3D14.4%5Bs%5D)