C: The gravitational force exerted from the planet on Moon A is two times smaller than the gravitational force exerted from the planet on Moon B.
Answer:
ummmm sea urchents are mostly in some areas??
Explanation:
thats ny guess??
The weight is the force of gravity acting on the object. So assuming it’s on earth, 12(9.81) also, F(weight in Newtons)=mass( in kg)acceleration (due to gravity)
Isn't it "gravity" this would makes sense because grvaity difines weight
To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.
Torque in a body is defined as,
![\tau_l = F*d](https://tex.z-dn.net/?f=%5Ctau_l%20%3D%20F%2Ad)
And in angular movement like
![\tau_a = I*\alpha](https://tex.z-dn.net/?f=%5Ctau_a%20%3D%20I%2A%5Calpha)
Where,
F= Force
d= Distance
I = Inertia
Acceleration Angular
PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.
![\tau= F*cos(19)*d](https://tex.z-dn.net/?f=%5Ctau%3D%20F%2Acos%2819%29%2Ad)
On the other hand we have the speed data expressed in RPM, as well
![\omega_f = 10rpm = 10\frac{1rev}{1min}(\frac{1min}{60s})(\frac{2\pi rad}{1rev})](https://tex.z-dn.net/?f=%5Comega_f%20%3D%2010rpm%20%3D%2010%5Cfrac%7B1rev%7D%7B1min%7D%28%5Cfrac%7B1min%7D%7B60s%7D%29%28%5Cfrac%7B2%5Cpi%20rad%7D%7B1rev%7D%29)
![\omega_f = 1.0471rad/s](https://tex.z-dn.net/?f=%5Comega_f%20%3D%201.0471rad%2Fs)
Acceleration can be calculated by
![\alpha = \frac{\omega_f}{t}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5Comega_f%7D%7Bt%7D)
![\alpha = \frac{1.0471}{9}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1.0471%7D%7B9%7D)
![\alpha = 0.11rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%200.11rad%2Fs%5E2)
In the case of Inertia we know that it is equivalent to
![I = \frac{1}{2}mr^2 = \frac{1}{2}(750)*(2.3)^2](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmr%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28750%29%2A%282.3%29%5E2)
![I = 1983.75kg.m^2](https://tex.z-dn.net/?f=I%20%3D%201983.75kg.m%5E2)
Matching the two types of torque we have to,
![\tau_l=\tau_a](https://tex.z-dn.net/?f=%5Ctau_l%3D%5Ctau_a)
![Fd=I\alpha](https://tex.z-dn.net/?f=Fd%3DI%5Calpha)
![Fcos(19)*2.3=1983.75(0.11)](https://tex.z-dn.net/?f=Fcos%2819%29%2A2.3%3D1983.75%280.11%29)
![F=100.34N](https://tex.z-dn.net/?f=F%3D100.34N)
PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,
![W = \frac{1}{2}I\omega_f^2](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7B1%7D%7B2%7DI%5Comega_f%5E2)
![W= \frac{1}{2}(1983.75)(1.0471)^2](https://tex.z-dn.net/?f=W%3D%20%5Cfrac%7B1%7D%7B2%7D%281983.75%29%281.0471%29%5E2)
![W=1087.51J](https://tex.z-dn.net/?f=W%3D1087.51J)