Answer:
The value is 
Explanation:
From the question we are told that
The focal length of the objective is 
The focal length of the eyepiece is 
The tube length is 
Generally the magnitude of the overall magnification is mathematically represented as

Where
is the objective magnification which is mathematically represented as

=> 
=> 
is the eyepiece magnification which is mathematically evaluated as



So


I’m not going to church tomorrow or Friday I don’t want to go go back up
Answer:
3) Ep = 13243.5[J]
4) v = 17.15 [m/s]
Explanation:
3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.
m = mass = 90 [kg]
h = elevation = 15 [m]
Potential energy is defined as the product of mass by gravity by height.
![E_{p}=m*g*h\\E_{p}=90*9.81*15\\E_{p}=13243.5[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3Dm%2Ag%2Ah%5C%5CE_%7Bp%7D%3D90%2A9.81%2A15%5C%5CE_%7Bp%7D%3D13243.5%5BJ%5D)
This energy will be transformed into kinetic energy.
Ek = 13243.5 [J]
4) The velocity can be determined by defining the kinetic energy, as shown below.
![E_{k}=\frac{1}{2} *m*v^{2} \\v = \sqrt{\frac{2*E_{k} }{m} }\\ v= \sqrt{\frac{2*13243.5 }{90} }\\v=17.15[m/s]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%20%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B2%2AE_%7Bk%7D%20%7D%7Bm%7D%20%7D%5C%5C%20v%3D%20%5Csqrt%7B%5Cfrac%7B2%2A13243.5%20%7D%7B90%7D%20%7D%5C%5Cv%3D17.15%5Bm%2Fs%5D)
Answer:
angular velocity(ω) is the rate change of angular displacement.
ω=θ/t and it SI unit is rad/s
Explanation:
this is very similar with the definition of linear velocity (rate of change of displacement). it specifies the angular speed of an object and the axis about which the object is rotating.