Answer:

Explanation:
The strength of the gravitational field at the surface of a planet is given by
(1)
where
G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
For the Earth:

For the unknown planet,

Substituting into the eq.(1), we find the gravitational acceleration of planet X relative to that of the Earth:

And substituting g = 9.8 m/s^2,

Answer:
The value is 
Explanation:
From the question we are told that
The circuit resistance is
The feedback resistance is
The offset current is 
Generally the offset voltage is mathematically reparented as

=> 
=> 
Potential equals kenecric at the bottom so potential would also increas
Number 5 is A
number 6 is D
Answer:
ω = 1.83 rad/s clockwise
Explanation:
We are given:
I1 = 3.0kg.m2
ω1 = -5.4rad/s (clockwise being negative)
I2 = 1.3kg.m2
ω2 = 6.4rad/s (counterclockwise being positive)
By conservation of the momentum:
I1 * ω1 + I2 * ω2 = (I1 + I2) * ω
Solving for ω:

Since it is negative, the direction is clockwise.