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Yuliya22 [10]
3 years ago
7

By standard convention, both the electric potential and the the electric potential energy between two charges is taken to be zer

o in what configuration?
Physics
1 answer:
shusha [124]3 years ago
6 0

Answer: at when distance r = infinity.

Explanation: The formulae for the electric potential of an electric charge to an arbitrary point is given by the formulae below

V = q/4πεr

V = electric potential (volts)

q = magnitude of electric charge

ε = permittivity of free space

r = distance between arbitrary point and charge.

In the equation above, it can be seen that only electric potential (v) and distance (r) is a variable, and there is an inverse relationship between them (an increase in one leads to a decrease in the other)

Thus to have zero value of electric potential (v= 0) we have to have the largest value of r ( r = infinity).

Same goes for electric potential energy between two charges, the formulae is given below as

W = q1 *q2/4πεr

W= electric potential energy

q1 = magnitude of first charge.

q2 = magnitude of second charge

ε = permittivity of free space

r = distance between arbitrary point and charge.

Also, all values are constant aside from electric potential energy (w) and distance (r) which have an inverse relationship.

Thus to have zero value of electric potential energy (w =0), we have to get an infinite value of distance ( r =infinity)

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If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.
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Answer:

400ft.    32ft/s       -32ft/s

Explanation:

In reality the gravitational acceleration is 9.81 so the quadratic coefficient of the function should be 9.81/2

Anyway for the sake of assumtion let us takes=160t-16t^2

 

ds/dt=160-32t=0

 

t=160/32= 5 seconds.

s=160*160/32-16*(160/32)^2= 400 mts

 

 

s=384 mts

160t-16t^2=384

i.e

16t^2-160t+384=0

 

t^2-10t+24=0

(t-6)(t-4)=0

t=[4,6]

we have to take t=4 because it is all the up i.e <5

 

velocity =v=ds/dt=160-32t

 

v=160-32*4=32 ft/sec still going up

 

for all the way down take t=6 whuch is >5

 

v=160-6*32=-32 ft/sec (falling down!!!)

6 0
3 years ago
If an object was accelerating at 10 m/s2, and a mass of 1 kg, what was size of the force acting on the object?
ryzh [129]

Answer:

10 N

Explanation:

f = ma

= 10m/s^2 * 1 kg

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Answer:

Explanation:

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