Answer:
<h2> 4kg</h2>
Explanation:
Step one:
given
length of rod=2m
mass of object 1 m1=1kg
let the unknown mass be x
center of mass<em> c.m</em>= 1.6m
hence 1kg is 1.6m from the <em>c.m</em>
and x is 0.4m from the <em>c.m</em>
Taking moment about the <em>c.m</em>
<em>clockwise moment equals anticlockwise moments</em>
1*1.6=x*0.4
1.6=0.4x
divide both sides by 0.4 we have
x=1.6/0.4
x=4kg
The mass of the other object is 4kg
The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v
<u>Explanation</u>:
Given data,
charge = 5.5 x 10¹² C
k =9.00 x 10⁹
The electric potential V of a point charge can found by,
V= kQ / r
Assuming, r=5.00×10⁻² m
V= 5.5 x 10⁻¹²C x 9.00 x 10⁹ / 5.00×10⁻² m
V= 49.5 x 10⁻³/ 5.00×10⁻²
Electric potential V= 0.099 x 10⁻¹v
Answer : 413.44N
Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .
- From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
- The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
- When a elevator accelerates down , the weight recorded is less than the actual weight .
From the Free body diagram ,
- Mass of the man = 64.2 kg
The answer is unbalanced because the numbers 100n and 300n don’t equal each other. If the answer was balanced the numbers would have to equal each other. Also, the answer net force would only be the answer if we were adding the numbers. I hope this helped you! :)