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Vikki [24]
3 years ago
14

Two tiny beads are 25 cm apart with no other charged objects or fields present. Bead A has a net charge of magnitude 10 nC and b

ead B has a net charge of magnitude 1 nC. Which one of the following statements is true about the magnitudes of the electric forces on these beads?A. The force on A is 100 times the force on B.B. The force on B is 10 times the force on A.C. The force on B is 100 times the force on A.D. The force on A is 10 times the force on B.E. The force on A is exactly equal to the force on B.
Physics
1 answer:
Alona [7]3 years ago
6 0

Answer:

E. The force on A is exactly equal to the force on B.

Explanation:

The force between two charges is given by

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

where

q_1 = Charge on particle 1

q_2 = Charge on particle 2

r = Distance between the charges

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

F=\dfrac{8.99\times 10^9\times 10\times 10^{-9}\times 1\times 10^{-9}}{(25\times 10^{-2})^2}\\\Rightarrow F=0.0000014384\ N

This force will be exerted on both the charges equally.

So, The force on A is exactly equal to the force on B

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wo ships, one 200200 metres in length and the other 100100 metres in length, travel at constant but different speeds. When trave
shutvik [7]

Answer:

The speed of the faster ship is 22.5 m/s

Explanation:

The length of the ships are;

Ship 1 = 200 m

Ship 2 = 100 m

The time it takes for the ships to completely cross each other when travelling in opposite directions = 10 seconds

The time it takes both ships to cross each other when travelling in the same direction = 25 seconds

Let x represent the speed of the first ship, ship 1, and y represent the speed of the second ship, ship, 2, we have;

(x + y) × 10 = 200 + 100 = 300

10·x + 10·y = 300...(1)

(x - y) × 20 = 200 + 100 = 300

20·x - 20·y = 300...(2)

Multiply equation (1) by 2, to get;

(x + y) × 10 × 2 = 300 × 2

20·x + 20·y = 600...(3)

Adding equation (1) to equation (3) gives;

20·x + 20·y + 20·x - 20·y = 600 + 300

40·x = 900

x = 900/40 = 22.5

x = 22.5

The speed of the first ship, ship 1 = x = 22.5 m/s

From equation (1), we have;

10·x + 10·y = 300

y = (300 - 10·x)/10 = (300 - 10×22.5)/10 = 7.5

y = 7.5

The speed of the second ship, ship 2 = y = 7.5 m/s

Therefore, the faster ship is ship 1 with a speed of 22.5 m/s

7 0
2 years ago
Which of these is an example of qualitative observation?
Zanzabum
Iron nails are attracted to the magnet
4 0
2 years ago
Read 2 more answers
The force of repulsion between two like-charged particles will increase if
svlad2 [7]
If they're brought closer together
5 0
2 years ago
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Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up
Westkost [7]

Answer:

24 N

Explanation:

m = mass of the cube = 6.0 kg

Consider the three cubes together as one.

M = mass of the three cubes together = 3 m = 3 (6.0) = 18 kg

a = acceleration of the combination = 2 ms⁻²

F = Force applied on the combination

Using Newton's second law

F = ma = (18) (2) = 36 N

F_{L} = Force by the left cube on the middle cube

Consider the forces acting on left cube, from the force diagram, we have

F - F_{L} = ma \\36 - F_{L} = (6) (2)\\F_{L} = 24 N

4 0
2 years ago
A jogger hears a car alarm and decides to investigate. While running toward the car, she hears an alarm frequency of 872.10 Hz.
Nata [24]

Answer:

v = 4.18 m/s

Explanation:

given,

frequency of the alarm = 872.10 Hz

after passing car frequency she hear = 851.10 Hz

Speed of sound = 343 m/s

speed of the jogger = ?

speed of the

v_f = \dfrac{872.10-851.10}{2}

v_f =10.5\ Hz

v_o = 872.1 - 10.5

V_0 = 861.6\ Hz

The speed of jogger

v = \dfrac{v_1 \times 343}{v_0}-343

v = \dfrac{872.1 \times 343}{861.6}-343

v = 4.18 m/s

5 0
2 years ago
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