Answer:
- The gravitational force by the Earth on the object, and by the object on the Earth is
- = 6.674×10−11 m3⋅kg−1⋅s−2 × 6 × 10^24 kg × 44.5 kg/(6.4 × 10^6 m)²
<u>Please note that the ration between the gravitation force 435 and the mass 44.5</u>
- should be gravitational acceleration
- I attribute the discrepancy between 9.78 and the usual 9.81 to rounding off in the
- Earth's weight and radius.
The mass of the Moon is M / 81.3.
The radius of the Moon is R × 0.27.
The gravitational force on the moon would be
G(M/81.3)m/(R×0.27)² = 0.17×GMm/R²
The gravitational force on the moon is smaller by the factor of about 0.17.
Answer:
DL = 1.5*10^-4[m]
Explanation:
First we will determine the initial values of the problem, in this way we have:
F = 60000[N]
L = 4 [m]
A = 0.008 [m^2]
DL = distance of the beam compressed along its length [m]
With the following equation we can find DL
![\frac{F}{A} = Y*\frac{DL}{L} \\where:\\Y = young's modulus = 2*10^{11} [Pa]\\DL=\frac{F*L}{Y*A} \\DL=\frac{60000*4}{2*10^{11} *0.008} \\DL= 1.5*10^{-4} [m]](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BA%7D%20%3D%20Y%2A%5Cfrac%7BDL%7D%7BL%7D%20%5C%5Cwhere%3A%5C%5CY%20%3D%20young%27s%20modulus%20%3D%202%2A10%5E%7B11%7D%20%5BPa%5D%5C%5CDL%3D%5Cfrac%7BF%2AL%7D%7BY%2AA%7D%20%5C%5CDL%3D%5Cfrac%7B60000%2A4%7D%7B2%2A10%5E%7B11%7D%20%2A0.008%7D%20%5C%5CDL%3D%201.5%2A10%5E%7B-4%7D%20%5Bm%5D)
Note: The question should be related with the distance, not with the diameter, since the diameter can be found very easily using the equation for a circular area.
![A=\frac{\pi}{4} *D^{2} \\D = \sqrt{\frac{A*4}{\pi} } \\D = \sqrt{\frac{0.008*4}{\\pi } \\\\D = 0.1[m]](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2AD%5E%7B2%7D%20%5C%5CD%20%3D%20%5Csqrt%7B%5Cfrac%7BA%2A4%7D%7B%5Cpi%7D%20%7D%20%5C%5CD%20%3D%20%20%5Csqrt%7B%5Cfrac%7B0.008%2A4%7D%7B%5C%5Cpi%20%7D%20%5C%5C%5C%5CD%20%3D%200.1%5Bm%5D)
Answer:
All trenches must have access/egress - A.
Rock 3, because it has the most mass while still moving at the same speed, it is kinetic because there is movement and energy being used
In a real system of levers, wheel or pulleys, the AMA (actual mechanical advantage) is less than the IMA (ideal mechanical advantage) because of the presence of friction.
In fact, the IMA and the AMA of a machine are defined as the ratio between the output force (the load) and the input force (the effort):

however, the difference is that the IMA does not take into account the presence of frictions, while the AMA does. As a result, the output force in the AMA is less than the output force in the IMA (because some energy is dissipated due to friction), and the AMA is less than the IMA.