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

=> 
=> 
Answer:
It makes it lighter when its closer and heavier when its farther way.
Explanation:
Answer:
Increase 9 times
Explanation:
We have Newton formula for attraction force between 2 objects with mass and a distance between them:

where
is the gravitational constant.
is the masses of the 2 objects. and R is the distance between them.
Since the force is inversely proportional to the distance squared, if it is reduced by 3 times, the gravitational force between them would increase by
times
Answer:

Explanation:
The total force on the particle is given by

Then, by replacing we have:
![q\vec{v}\ X \vec{B}=q[7\hat{k}-9\hat{j}-\hat{k}]\\\\q\vec{E}=q[5\hat{i}-\hat{j}-2\hat{k}]\\\\\vec{F}=(9.61*10^{-19}C)[(7+9)\hat{i}+(-9-1)\hat{j}+(-1-2)\hat{k}]\\\\\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N](https://tex.z-dn.net/?f=q%5Cvec%7Bv%7D%5C%20X%20%5Cvec%7BB%7D%3Dq%5B7%5Chat%7Bk%7D-9%5Chat%7Bj%7D-%5Chat%7Bk%7D%5D%5C%5C%5C%5Cq%5Cvec%7BE%7D%3Dq%5B5%5Chat%7Bi%7D-%5Chat%7Bj%7D-2%5Chat%7Bk%7D%5D%5C%5C%5C%5C%5Cvec%7BF%7D%3D%289.61%2A10%5E%7B-19%7DC%29%5B%287%2B9%29%5Chat%7Bi%7D%2B%28-9-1%29%5Chat%7Bj%7D%2B%28-1-2%29%5Chat%7Bk%7D%5D%5C%5C%5C%5C%5Cvec%7BF%7D%3D%281.537%2A10%5E%7B-17%7D%5Chat%7Bi%7D-9.61%2A10%5E%7B-19%7D%5Chat%7Bj%7D-2.883%2A10%5E%7B-18%7D%5Chat%7Bk%7D%29N)
where the cross product can be made with the determinant method.
Hope this helps!!
Energy to lift something =
(mass of the object) x (gravity) x (height of the lift).
BUT ...
This simple formula only works if you use the right units.
Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters
For this question . . .
Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms
Gravity (on Earth) = 9.8 m/second²
Height = 500 cm = 5.0 meters
So we have ...
Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)
= 2,696,925 joules .
That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
hour, or burning a 100 watt light bulb for about 7-1/2 hours.
The reason is the large mass that's being lifted.
On Earth, that much mass weighs about 61 tons.