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stiv31 [10]
3 years ago
14

Two metal spheres are suspended from strings. The

Physics
1 answer:
nikdorinn [45]3 years ago
3 0

Answer:

B. A repulsive force of 8.0*10^3 N.

Explanation:

As we know by Coulomb's law that the electrostatic force between two charges is given as

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 = -2 \times 10^2 C

q_2 = -4 \times 10^{-8} C

r = 3.0 m

now we have

F = \frac{(9 \times 10^9)(2 \times 10^2)(4\times 10^{-8})}{3^2}

F = 8000 N

since both charges are similar charges so they will repel each other by the force we calculated above so correct answer will be

B. A repulsive force of 8.0*10^3 N.

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Answer:

14.3°C

Explanation:

Find the ratio of 10°C : 700ml then use the same ratio to 1000ml.

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<span>-- the product of the net charges on the objects;. -- the distance between the centers of their net charges. (Pretty much identical to the formula for gravitational force)</span>
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• Usain Bolt can run at 10 m/s. If he runs for about 20 seconds, how far around the race track did he go?
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200metters

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Difference between discrete and continuous charge distribution?
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3 years ago
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The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
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