Consider a collection of charges in a given region and suppose all other charges are distant and have a negligible effect. Furth
1 answer:
Answer:
E. Some charges in the region are positive, and some are negative.
Explanation:
Electric potential is given as;
where;
W is the work done in moving a charge between two points which have a difference in potential
Q is quantity of charge in the given region
If the electric potential at a given point in the region is zero, then sum of the charges in the given region must be equal to zero. For the charges to sum to zero, some will be positive while some will be negative,.
Therefore, the correct statement in the given options is "E"
E. Some charges in the region are positive, and some are negative.
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(a) 30.9 m
Let's analyze the motion of the parachutist. Its vertical position above the ground is given by
where
h = 45 m is the initial height
u = 10 m/s is the initial velocity (upward)
t is the time
g = -9.8 m/s^2 is the acceleration of gravity (downward)
Substituting t=3 s , we find the height of the parachutist when it opens the parachute:
(b) 44.1 m
Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by
y = h + ut
where
h = 45 m is the initial height
u = 10 m/s is the initial velocity (upward)
t is the time
Substituting t = 3 s, we find
y = 45 m + (10 m/s)(3 s) = 75 m
So the distance between the balloon and the parachutist after 3 s is
d = 75 m - 30.9 m = 44.1 m
(c) 8.2 m/s downward
The velocity of the parachutist at the moment he opens the parachute is:
v = u +gt
where
u = 10 m/s is the initial velocity (upward)
t is the time
g = -9.8 m/s^2 is the acceleration of gravity (downward)
Substituting t = 3 s,
v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s
where the negative sign means it is downward
After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of
a =+5 m/s^2
where we put a positive sign since this time the acceleration is upward.
The total distance he still has to cover till the ground is
d = 30.9 m
So we can find the final velocity by using
where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:
a = -5 m/s^2
Solving for v,
(d) 5.24 s
We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using
where
a = -5 m/s^2 is the deceleration
v = 8.2 m/s is the final velocity
u = 19.4 m/s is the initial velocity
t is the time
Solving for t, we find
And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is
t = 3 s + 2.24 s = 5.24 s
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