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3 years ago

Question 5

Physics

1 answer:

MAXImum [283]3 years ago

5 0

Answer:

Explanation:

$m =800kg\\v = 15\\\\K.E = \frac{1}{2}mv^2\\ K.E= \frac{1}{2} \times 800\times 15^2\\= 400 \times 225\\= 90000 joules\\= 90 kilojoules$

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What is the potential energy of the system composed of the three charges q1, q3, and q4, when q1 is at point R? Define the poten

Ad libitum [116K]

Answer:

Incomplete question check attachment for complete question

Also it is given that

q1=-0.7uC

q4=-1.7uC

q3=-1.7uC

Also the distance are given as

a=2.2cm=0.022m

d2=3.6cm=0.036m

Explanation:

The potential energy due to point R is given as

The potential energy due to charge q1 and q3 plus the potential energy due to charge q4 and q1 plus the potential energy due to charge q3 and q4

So, let take it one after the other

Potential energy is give as

P.E=kq1q2/r

Therefore,

Potential energy due to charge q1 and q3

U¹³=kq1q3/r

To get the distance between charge q1 and q3, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

k is a constant =9×10^9Nm²/C²

Then,

U¹³=kq1q3/r

U¹³=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹³=0.254J

Potential energy due to charge q1 and q4

U¹⁴=kq1q4/r

To get the distance between charge q1 and q4, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

k is a constant =9×10^9Nm²/C²

Then,

U¹⁴=kq1q4/r

U¹⁴=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹⁴=0.254J

Potential energy due to charge q3 and q4

U³⁴=kq3q4/r

r=2a=2×0.022=0.044m

k is a constant =9×10^9Nm²/C²

Then,

U³⁴=kq3q4/r

U³⁴=9×10^9×1.7×10^-6×1.7×10^-6/0.044

U¹⁴=0.591J

Then, the total energy is

U= U¹³+ U¹⁴ + U³⁴

U=0.254+0.254+0.591

U=1.099J

Then also, the potential energy is zero because at infinity both U¹³ and U¹⁴ will have infinite potential because their distance apart will be infinite.

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3 years ago