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Fynjy0 [20]
3 years ago
9

​In Figure 4.24, a current of 0.3 A flows through the conductor CD, and a charge of 4C passes through a cross-section AB of the

conductor. Find the time taken by the charge to pass through AB​
Physics
2 answers:
olga2289 [7]3 years ago
8 0

Answer:

13.33 seconds

Explanation:

I = Q/t

t = Q/I = 4/0.3 = 13.33 seconds

strojnjashka [21]3 years ago
7 0

This is your answer . I hope it will help you .

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Statements 1, 3, and 5 are true.

(A, C, and E)

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An athelete runs some distance before taking a long jump because : give answer
insens350 [35]
Hi Pupil Here's your answer ::



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3 0
3 years ago
Scientists want to place a telescope on the moon to improve their view of distant planets. The telescope weighs 200 pounds on Ea
Maru [420]

Answer:

33,02 lb

Explanation:

g_m ≈ 1,62 m/s2

g ≈ 9,81 m/s2

m = 200 lb

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7 0
3 years ago
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Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on
disa [49]

Answer:

V = 48 Volts

Explanation:

Since we know that electric potential is a scalar quantity

So here total potential of a point is sum of potential due to each charge

It is given as

V = V_1 + V_2 + V_3

here we have potential due to 50 nC placed at y = 6 m

V_1 = \frac{kQ}{r}

V_1 = \frac{(9\times 10^9)(50 \times 10^{-9})}{\sqrt{6^2 + 8^2}}

V_1 = 45 Volts

Now potential due to -80 nC charge placed at x = -4

V_2 = \frac{kQ}{r}

V_2 = \frac{(9\times 10^9)(-80 \times 10^{-9})}{12}

V_2 = -60 Volts

Now potential due to 70 nC placed at y = -6 m

V_3 = \frac{kQ}{r}

V_3 = \frac{(9\times 10^9)(70 \times 10^{-9})}{\sqrt{6^2 + 8^2}}

V_3 = 63 Volts

Now total potential at this point is given as

V = 45 - 60 + 63 = 48 Volts

3 0
3 years ago
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