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Lelu [443]
2 years ago
10

What is the electrical force between q2 and q3? recall that k = 8.99 × 109 n•meters squared over coulombs squared.. 1.0 × 1011 n

–1.1 × 1011 n –1.6 × 1011 n 1.8 × 1011 n
Physics
1 answer:
max2010maxim [7]2 years ago
6 0

The magnitude of the electrical force between q2 and q3 is given as a ratio between the product of their charges and the square of the distance of separation.

<h3>What is the magnitude of electrical forces between two charges?</h3>

The magnitude of the electrical force between two charges refers to the attractive or repulsive forces that exists between two charges separated by a given distance in an electric field.

The magnitude of the electrical force, F between the two charges q2 and q3 is given be my the formula below

F = \frac{K \times q_2 \times q_3}{d^{2}}

Therefore, the magnitude of the electrical force between q2 and q3 is given as a ratio between the product of their charges and the square of the distance of separation.

Learn more about electrical force at: brainly.com/question/17692887

#SPJ4

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Two sound waves, A and B, are traveling at the same speed. Wave A has a wavelength of 50 cm and a frequency of 7000 Hz. Wave B h
vekshin1

Answer:

D. 100 cm

Explanation:

The speed of a wave is the wavelength times the frequency.

v = λf

Wave A and B have the same speed, so:

λf = λf

(50 cm) (7000 Hz) = λ (3500 Hz)

λ = 100 cm

6 0
3 years ago
A 2kg object is moving horizontally with a speed of 4m/s 2. How much net force is required t keep the object moving at this spee
dezoksy [38]
To continue moving at constant speed in a straight line requires NO net force. Zero. Nada. If there IS any net force on the object, then its speed or direction will change.
8 0
3 years ago
What is the magnetic field strength required to make a proton with a speed of 5.0 x 10^(5) m/s follow a circular path of radius
Nuetrik [128]

Answer:

Magnetic field strength required for this is 0.25 T

Explanation:

As we know that the proton moves in circular path in uniform magnetic field

so the radius of the path of the circle is given as

R = \frac{mv}{qB}

here we know that

q = 1.6 \times 10^{-19}C

m = 1.6 \times 10^{-27} kg

R = 0.0200 m

v = 5 \times 10^5 m/s

now we have

B = \frac{1.6 \times 10^{-27} (5 \times 10^5)}{(1.6 \times 10^{-19})(0.02)}

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8 0
4 years ago
The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an instant when the ma
Law Incorporation [45]

Answer:

The speed of the particle is 2.86 m/s

Explanation:

Given;

radius of the circular path, r = 2.0 m

tangential acceleration,  a_t = 4.4 m/s²

total magnitude of the acceleration, a = 6.0 m/s²

Total acceleration is the vector sum of  tangential acceleration and radial acceleration

a = \sqrt{a_c^2 + a_t^2}\\\\

where;

a_c is the radial acceleration

a = \sqrt{a_c^2 + a_t^2}\\\\a^2 = a_c^2 + a_t^2\\\\a_c^2 = a^2 -a_t^2\\\\a_c = \sqrt{a^2 -a_t^2}\\\\a_c = \sqrt{6.0^2 -4.4^2}\\\\a_c = \sqrt{16.64}\\\\a_c = 4.08 \ m/s^2

The radial acceleration relates to speed of particle in the following equations;

a_c = \frac{v^2}{r}

where;

v is the speed of the particle

v^2 = a_c r\\\\v= \sqrt{a_c r} \\\\v = \sqrt{4.08 *2}\\\\v = 2.86 \ m/s

Therefore, the speed of the particle is 2.86 m/s

6 0
4 years ago
A high diver of mass 74.0 kg jumps off a board 9.00 m above the water. If his downward motion is stopped 2.50 seconds after he e
stiv31 [10]

Answer:

1120 N

Explanation:

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v² = (0 m/s)² + 2 (-9.8 m/s²) (-9.00 m)

v = -13.3 m/s

Or it can be found with conservation of energy.

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × -9.8 m/s² × -9.00 m)

v = -13.3 m/s

Sum of forces on the diver after he hits the water:

∑F = ma

F − mg = m Δv/Δt

F − (74.0 kg) (9.8 m/s²) = (74.0 kg) (0 m/s − (-13.3 m/s)) / (2.50 s)

F = 1120 N

6 0
3 years ago
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