The potential energy of the block is A) 490 J
Explanation:
The potential energy of an object is the energy possessed by the object due to its position in the gravitational field.
It is calculated as follows:

where
m is the mass of the object
g is the acceleration due to gravity
h is the height of the object above the ground
For the block in this problem, we have:
m = 10 kg

h = 5 m
Therefore, its potential energy is:

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Answer:
<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>
Explanation:
The image is shown below.
The force on the particle with charge q due to each charge Q = 
we designate this force as N
Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.
Resolving the forces on the particle, we have
for the x-component
= N cosine 60° + (-N cosine 60°) = 0
for the y-component
= -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N
The above indicates that there is no resultant force in the x-axis, since it is equal to zero (
= 0).
The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.
<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>
The electrostatic force between two charges is inversely
proportional to the square of the distance between them.
So if you want to multiply the force by, say, ' Q ',
you need to multiply the distance by ( 1 / √Q ) .
We want to multiply the force by 16, so we need to
multiply the distance by ( 1 / √16 ) = ( 1 / 4 ) .
The distance should be changed to 1/4 of what it is now.
(i) The total capacitance for the circuit is 5 μF.
(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.
(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.
<h3>Total capacitance of the circuit</h3>
The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.
<h3>C1 and C2 are in series </h3>

<h3>C1 and C2 are parallel to C3</h3>

<h3>C(123) is series to C5 and C6</h3>

<h3>C7 and C8 are in series</h3>

<h3>Total capaciatnce of the circuit</h3>
Ct + C(78) = 2 μF + 3 μF = 5 μF
<h3 /><h3>Total charge stored in the circuit</h3>
The total charge stored in the capacitor is calculated as follows;
Q = CV
Q = (5 x 10⁻⁶) x (20)
Q = 1 x 10⁻⁴ C
<h3>Charge stored in 3μF capacitor</h3>
Q = (3 x 10⁻⁶) x (20)
Q = 6 x 10⁻⁶ C
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Answer:
Average speed is total distance divided by total time.
v = d / t