To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.
PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation
![g = \frac{GM}{(d-R_{CM})^2}](https://tex.z-dn.net/?f=g%20%3D%20%5Cfrac%7BGM%7D%7B%28d-R_%7BCM%7D%29%5E2%7D)
Where,
G = Gravitational Universal Constant
d = Distance
M = Mass
Radius earth center of mass
PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,
![g = \frac{GM}{(d-R_{CM})^2}](https://tex.z-dn.net/?f=g%20%3D%20%5Cfrac%7BGM%7D%7B%28d-R_%7BCM%7D%29%5E2%7D)
![g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}](https://tex.z-dn.net/?f=g%20%3D%20%5Cfrac%7B%286.67%2A10%5E%7B-11%7D%29%287.35%2A10%5E%7B22%7D%29%7D%7B%283.84%2A10%5E8-4700%2A10%5E3%29%5E2%7D)
![g = 3.4*10^{-5}m/s^2](https://tex.z-dn.net/?f=g%20%3D%203.4%2A10%5E%7B-5%7Dm%2Fs%5E2)
PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is
![\omega = \frac{2\pi}{T}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B2%5Cpi%7D%7BT%7D)
At the same time we have that centripetal acceleration is given as
![a_c = \omega^2 r](https://tex.z-dn.net/?f=a_c%20%3D%20%5Comega%5E2%20r)
Replacing
![a_c = (\frac{2\pi}{T})^2 r](https://tex.z-dn.net/?f=a_c%20%3D%20%28%5Cfrac%7B2%5Cpi%7D%7BT%7D%29%5E2%20r)
![a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)](https://tex.z-dn.net/?f=a_c%20%3D%20%28%5Cfrac%7B2%5Cpi%7D%7B26.3d%28%5Cfrac%7B86400s%7D%7B1days%7D%29%7D%29%5E2%20%284700%2A10%5E3m%29)
![a_c = 3.34*10^{-5}m/s^2](https://tex.z-dn.net/?f=a_c%20%3D%203.34%2A10%5E%7B-5%7Dm%2Fs%5E2)
Laser means Light Amplication by Stimulated Emission of Radiation.
Uses of laser
DNA sequencing instrument
Cutting and welding materials
Semiconducting chip manufacturing
Military devices
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Answer:
Anybody which is in state of rest ,will be in rest if we don't apply any external force ...
Answer:
a) 20 seconds
b) No.
Explanation:
t = Time taken for jet to stop
u = Initial velocity = 100 m/s (given in the question)
v = Final velocity = 0 (because the jet will stop at the end)
s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)
a = Acceleration = -5.00 m/s² (slowing down, so it is negative)
a) Equation of motion
![v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-100}{-5}\\\Rightarrow t=20\ s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%20t%3D%5Cfrac%7Bv-u%7D%7Ba%7D%5C%5C%5CRightarrow%20t%3D%5Cfrac%7B0-100%7D%7B-5%7D%5C%5C%5CRightarrow%20t%3D20%5C%20s)
The time required for the plane to slow down from the moment it touches the ground is 20 seconds.
![s=ut+\frac{1}{2}at^2\\\Rightarrow s=100\times 20+\frac{1}{2}\times -5\times 20^2\\\Rightarrow s=1000\ m](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%20s%3D100%5Ctimes%2020%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20-5%5Ctimes%2020%5E2%5C%5C%5CRightarrow%20s%3D1000%5C%20m)
The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.
Answer:
Negative
Explanation:
If the box is heading right in the positive direction, the work will be negative. The spring has an opposite force to that of the box.
Hope this helped. :)