Answer:
c. V = k Q1 * Q2 / R1 potential energy of Q1 and Q2 separated by R
V2 / V1 = (R1 / R2) = 1/4
V2 = V1 / 4
Question: A loader sack of total mass
is l000 grams falls down from
the floor of a lorry 200 cm high
Calculate the workdone by the
gravity of the load.
Answer:
19.6 Joules
Explanation:
Applying
W = mgh........................ Equation 1
Where W = Workdone by gravity on the load, m = mass of the loader sack, h = height, g = acceleration due to gravity
From the question,
Given: m = 1000 grams = (1000/1000) kilogram = 1 kg, h = 200 cm = 2 m
Constant: g = 9.8 m/s²
Substitute these values into equation 1
W = (1×2×9.8)
W = 19.6 Joules
Hence the work done by gravity on the load is 19.6 Joules
The change in velocity is 10 mi/h (4.47 m/s)
Explanation:
The change in velocity of the motorcyclist is given by

where
v is the final velocity
u is the initial velocity
In this problem, we have
u = 0 (the motorbike starts from rest)
v = 10 mi/h
Therefore, the change in velocity is

And keeping in mind that
1 mile = 1609 m
1 h = 3600 s
We can convert it into m/s:

Learn more about velocity:
brainly.com/question/5248528
#LearnwithBrainly
E = <u>kQ</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
(r + h)²
where,
k = 9 × 10^9Nm²C^-2
Q = total charge, 300uC = 300 × 10^ -6C
r = 8 × 10^ -2m
h = 16 × 10^ -2m
then,
E = <u>9</u><u>e</u><u>9</u><u> </u><u>*</u><u> </u><u>3</u><u>0</u><u>0</u><u>e</u><u>^</u><u>-</u><u>6</u><u> </u><u> </u><u> </u><u> </u>
(8e^-2 + 16e^-2)²
E = 4687500N/C
the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period
Thank you