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tangare [24]
2 years ago
13

An electric charge of 1 microcoulomb is affected by an electric field of intensity 1000 newtons per coulomb at a point, then thi

s point is a meter away from it by a distance
Physics
1 answer:
Alchen [17]2 years ago
8 0

Answer:

r = 3 m

Explanation:

Given that,

Charge, q=1\ mu C

Electric field at a point, E=1000\ N/C

We need to find the distance away from it. We know that electric field at point r is given by :

E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 10^{-6}}{1000}} \\\\r=3\ m

So, the required distance is 3 m.

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When the flashlight is in the air and the refracted ray enters the water, how does the angle of refraction compare with the angl
tigry1 [53]

Answer:

The refracted angle will be less than the angle of incidence and the speed of light is slower in the new medium (which is water) and is closer to the normal.

Try remembering this by using FST SFA (Fast Sofa)

                                             Fast -> Slow = Towards the normal

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3 years ago
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3 years ago
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Give an example of a compound machine. Explain how at least two simple machines are part of this complex machine.
11Alexandr11 [23.1K]

Answer:

Bicycle

Explanation:

A compound machine is a machine which is a combination of simple machines.

Simple machines are like the pulley, inclined plane or a screw.

Suppose a bicycle is considered, it has more than one simple machine combined together, for it to work. Wheel and axle is one of them and the beam which is pivoted at a fixed hinge is another simple machine in it.

The pedals of the bicycle function as the lever.

6 0
3 years ago
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener
defon

Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s

(a) the force constant of the spring

\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m

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3 years ago
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