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kogti [31]
1 year ago
7

A stationary charge is located between the poles of a horseshoe magnet. Is a magnetic force excerted of the charge? why?

Physics
1 answer:
vredina [299]1 year ago
7 0

A stationary charge is located between the poles of a horseshoe magnet. The magnetic force exerted by the charge is zero.

<h3>What is charge?</h3>

Charge is the physical property of matter which cause a particle to attract or repel when placed in its field.

A stationary charged particle does not interact with a static magnetic field. A charge placed in a magnetic field experiences a magnetic force. There will be no magnetic force acting on a stationary charge. The charge must be moving in order to have magnetic force on it.

Thus, the magnetic force exerted by the charge is zero.

Learn more about charge.

brainly.com/question/19886264

#SPJ4

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Why does changing the shape of an object have no effect on the density?
iren [92.7K]
Because the object is still made of the same material 
Density is not affected by the weight and shape of an object its affected by how concentrated the atoms are in a given volume 
7 0
3 years ago
You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation
Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, \omega = 25.13 rad/s

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = \frac{1}{T}

f = \frac{1}{0.25}

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

\omega = 2\pi \times f

\omega = 2\pi \times 4

\omega = 25.13 rad/s

5 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
What best describes the motion of the medium particles in a longitudinal wave? The medium does not move. The medium moves in all
Citrus2011 [14]

In the motion of the medium particles in a longitudinal wave, the medium vibrates parallel to the direction of the wave.

<h3>What is a longitudinal wave?</h3>

A longitudinal wave is a wave that is transversing along the length. When the displacement of medium and travel of wave is the same in that condition wave is known as the longitudinal wave.

It requires some medium to travel. A mechanical and sound wave is an example of a longitudinal wave.

Hence in the motion of the medium particles in a longitudinal wave, the medium vibrates parallel to the direction of the wave.

To learn more about the longitudinal wave refer to the link;

brainly.com/question/8497711

3 0
2 years ago
A flat coil having 160 turns, each with an area of 0.20 m 2, is placed with the plane of its area perpendicular to a magnetic fi
S_A_V [24]

Answer:

10.2 Watt

Explanation:

N  = number of turns in flat coil = 160

A  = area = 0.20 m²

B₀= initial magnetic field = 0.40 T

B  = final magnetic field = - 0.40 T

Change in magnetic field is given as

ΔB = B - B₀ = - 0.40 - 0.40 = - 0.80 T

t  = time taken for the magnetic field to change = 2.0 s

Induced emf is given as

E = \frac{- N A \Delta B}{t}

E = \frac{- (160) (0.20) (- 0.80)}{2}

E = 12.8 volts

R = Resistance of the coil = 16 Ω

Power is given as

P = \frac{E^{2}}{R}

P = \frac{(12.8)^{2}}{16}

P = 10.2 Watt

6 0
3 years ago
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