The answer would be b 0.84
To solve this problem we will apply Newton's second law, which indicates that the force is equivalent to the product between mass and acceleration, so
![F = ma](https://tex.z-dn.net/?f=F%20%3D%20ma)
Here,
F= Force
m = Mass
a = Acceleration
Rearranging to find the mass we have,
![m = \frac{F}{a}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7BF%7D%7Ba%7D)
The value of the acceleration is
![a = 18miles/hour^2 (\frac{0.00012417m/s^2}{1 miles/hour^2})](https://tex.z-dn.net/?f=a%20%3D%2018miles%2Fhour%5E2%20%28%5Cfrac%7B0.00012417m%2Fs%5E2%7D%7B1%20miles%2Fhour%5E2%7D%29)
![a = 0.002235m/s^2](https://tex.z-dn.net/?f=a%20%3D%200.002235m%2Fs%5E2)
Replacing to find the mass,
![m = \frac{53kN}{0.002235m/s^2}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B53kN%7D%7B0.002235m%2Fs%5E2%7D)
![m = \frac{53*10^{-3}}{0.002235}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B53%2A10%5E%7B-3%7D%7D%7B0.002235%7D)
![m = 23.71kg](https://tex.z-dn.net/?f=m%20%3D%2023.71kg)
Now in ponds this value is
![m = 23.71kg(\frac{2.205lb}{1kg})](https://tex.z-dn.net/?f=m%20%3D%2023.71kg%28%5Cfrac%7B2.205lb%7D%7B1kg%7D%29)
![m=52.8 lb](https://tex.z-dn.net/?f=m%3D52.8%20lb)
Therefore the mass of the spacecraft is 52.8lb
Explanation:
From a new Moon, when the Moon is situated in orbit between the Earth and the Sun with its dark side in shadow facing towards us, the amount of lunar surface visible from the Earth increases through waxing crescent, first quarter and waxing gibbous phases until it becomes full.
Or
The new moon is the phase that is invisible to us here on Earth because the moon is between the earth and the sun, and its illuminated side is facing away from us. A solar eclipse occurs when the moon moves in front of the sun, blocking it from our view on Earth.
Answer:
10.93 rad/s
Explanation:
If we treat the student as a point mass, her moment of inertia at the rim is
![I_r = mr^2 = 67.8*3.7^2 = 928.182 kgm^2](https://tex.z-dn.net/?f=I_r%20%3D%20mr%5E2%20%3D%2067.8%2A3.7%5E2%20%3D%20928.182%20kgm%5E2)
So the system moment of inertia when she's at the rim is:
![I_1 = I_d + I_r = 274 + 928.182 = 1202.182 kgm^2](https://tex.z-dn.net/?f=I_1%20%3D%20I_d%20%2B%20I_r%20%3D%20274%20%2B%20928.182%20%3D%201202.182%20kgm%5E2)
Similarly, we can calculate the system moment of inertia when she's at 0.456 m from the center
![I_2 = I_d + 67.8*0.456^2 = 274 + 14.1 = 288.1 kgm^2](https://tex.z-dn.net/?f=I_2%20%3D%20I_d%20%2B%2067.8%2A0.456%5E2%20%3D%20274%20%2B%2014.1%20%3D%20288.1%20kgm%5E2)
We can apply the law of angular momentum conservation to calculate the post angular speed when she's 0.456m from the center:
![I_1\omega_1 = I_2\omega_2](https://tex.z-dn.net/?f=I_1%5Comega_1%20%3D%20I_2%5Comega_2)
![\omega_2 = \omega_1\frac{I_1}{I_2} = 2.62*\frac{1202.182}{288.1} = 10.93 rad/s](https://tex.z-dn.net/?f=%5Comega_2%20%3D%20%5Comega_1%5Cfrac%7BI_1%7D%7BI_2%7D%20%3D%202.62%2A%5Cfrac%7B1202.182%7D%7B288.1%7D%20%3D%2010.93%20rad%2Fs)