Answer:
Wmoon = 131 [N]
Explanation:
We know that the weight of a body is equal to the product of mass by gravitational acceleration.
Since we are told that the gravitational acceleration of the moon is equal to one-sixth of the acceleration of Earth's gravitation. Then we must multiply the value of Earth's gravitation by one-sixth.
![w_{moon}=\frac{1}{6} *m*g\\w_{moon}=\frac{1}{6} *80*9.81\\w_{moon}=130.8 [N] = 131 [N]](https://tex.z-dn.net/?f=w_%7Bmoon%7D%3D%5Cfrac%7B1%7D%7B6%7D%20%2Am%2Ag%5C%5Cw_%7Bmoon%7D%3D%5Cfrac%7B1%7D%7B6%7D%20%2A80%2A9.81%5C%5Cw_%7Bmoon%7D%3D130.8%20%5BN%5D%20%3D%20131%20%5BN%5D)
Answer:
150 million kilometres
Explanation:
The astronomical unit (symbol: au, or AU or AU) is a unit of length, roughly the distance from Earth to the Sun and equal to 150 million kilometres (93 million miles) or 8.3 light minutes.
Answer:
the magnitude of the velocity of one particle relative to the other is 0.9988c
Explanation:
Given the data in the question;
Velocities of the two particles = 0.9520c
Using Lorentz transformation
Let relative velocity be W, so
v
= ( u + v ) / ( 1 + ( uv / c²) )
since each particle travels with the same speed,
u = v
so
v
= ( u + u ) / ( 1 + ( u×u / c²) )
v
= 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )
we substitute
v
= 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )
v
= 1.904c / ( 1 + 0.906304 )
v
= 1.904c / 1.906304
v
= 0.9988c
Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c
Explanation:
It is given that,
Speed, v₁ = 7.7 m/s
We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :




So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.