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3 years ago

13

Gaussian surfaces A and B enclose the same positive point charge. The area of surface A is two times larger than that of surface

B. How does the total electric flux through the two surfaces compare?

1 answer:

svlad2 [7]3 years ago

6 0

Answer:

The total electric flux through the two surfaces is equal.

Explanation:

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A 58.72 kg person holding a steel ball stands motionless on a frozen lake.

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Answer:16.44 kg

Explanation:

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3 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.77 μC. A conducting spherical shell of inner radius 4.00 cm and ou

madreJ [45]

Answer:

Part a)

E = 0

Part b)

$E = 6.77 \times 10^7 N/C$

Part c)

Electric field inside the conductor is again zero

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Part d)

$E = 8.52 \times 10^6 N/C$

Explanation:

Part a)

conducting sphere is of radius

R = 2 cm

so electric field inside any conductor is always zero

So electric field at r = 1 cm

E = 0

Part b)

Now at r = 3 cm

By Gauss law

$E = \frac{kq}{r^2}$

$E = \frac{(9\times 10^9)(6.77 \muC)}{0.03^2}$

$E = 6.77 \times 10^7 N/C$

Part c)

Again when we use r = 4.50 cm

then we will have

Electric field inside the conductor is again zero

$E = 0$

Part d)

Now at r = 7 cm

again by Gauss law

$E = \frac{kQ}{r^2}$

$E = \frac{(9\times 10^9)(6.77\mu C - 2.13\mu C)}{0.07^2}$

$E = 8.52 \times 10^6 N/C$

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3 years ago

I need a short answer ?

mars1129 [50]

Answer:

Explanation:

7a) t = d/v = 100/45cos14.5 = 2.29533...= 2.30 s

7b) h = ½(9.81)(2.29533/2)² = 6.46056... = 6.45 m

or

h = (45sin14.5)² / (2(9.81)) = 6.47 m

which rounds to the same 6.5 m when limiting to the two significant digits of the initial velocity.

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3 years ago

The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud

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<h2>The distance between students is 2.46 m</h2>

Explanation:

The force of attraction due to Newton's gravitation law is

F = $\frac{Gm_1m_2}{r^2}$

Here G is the gravitational constant

m₁ is the mass of one student

m₂ is the mass of second student .

and r is the distance between them

Thus r = $\sqrt{\frac{Gm_1m_2}{F} }$

If we substitute the values in the above equation

r = $\sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }$

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