Answer:
a) m=20000Kg
b) v=0.214m/s
Explanation:
We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.
For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, 
, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as 
, and the mass of the first and second coals as 
and 
respectively
We start with the transition between parts A and B, so we have:
Which means
And since we want the mass of the first coal thrown () we do:
Substituting values we obtain
For the transition between parts B and C, we can write:
Which means
Since we want the new final speed of the car (
) we do:
Substituting values we obtain
Answer: Workdone293.02KJ
Explanation: The equation to use to calculate Workdone = Change in KE + Change in PE
Assuming velocity is constant,KE becomes 0
Workdone= Change in PE=mg
W=92×9.8×325=293.02KJ
The strength of the electric field will be 215760 NC⁻¹. The concept of the electric field strength is used in the problem.
<h3>What is electric file strength?</h3>
The electric field strength is defined as the ratio of electric force and charge. Its unit is NC⁻¹.
The given data in the problem is;
E is the Electric Field Strength
k is the Colomb's constant = 8.99 x 10⁹ N.m²/C²
q is the magnitude of charge = 6 μC = 6 x 10⁻⁶ C
r is the distance = 0.5 m
The electric field strength is given by the formula;
Hence the strength of the electric field will be 215760 NC⁻¹.
To learn more about the electric field strength refer to the link;
brainly.com/question/4264413
Volume of cube = Length x width x height
= 1 x 1 x 1
= 1 cm cubed.
If this cube is dropped into 28.0 ML of water the new volume reading onthe graduated cylinder = 1 + 28
= 29 mL
