PHYSICS COLLEGE

Why Coulomb force is called "Mutual Force"???????????????????????????????????????

Answers

Answer 1

Answer:

Answer

The Columb's law is the same as Gravitational law

Explanation

As we see the formula of both Coulomb and Gravitational Law,

$F_g = GM_1M_2/R_2$ (1)

$F_c = kq_1q_2/r_2$ (2)

The masses (M) in formula (1) experiencing the force of gravitational pull with each other which varies with changing the distance. In the formula (2), the charges also are felling the forces on each other which varies with distance. The charges and masses are just like the objects which are experiencing the forces which have a common factor as distance. The gravitational force is also called the mutual forces.

Related Questions

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The New York Wheel is the world's largest Ferris wheel. It's 183 meters in diameter and rotates once every 37.3 min.1. Find the magnitude of the average velocity at the wheel's rim, over a 7.40-
min interval.

2.Find the magnitude of the average acceleration at the wheel's rim, over a 7.40-
min interval.

Answers

Answer:

Velocity =0.241 m/s

Acceleration = 7.21e-4 m/s²

Explanation:

The wheel travels through

Θ = (7.40/37.3)*360º = 71.42º

and so the length of the line segment connecting the initial and final position is

L = 2*L*sin(Θ/2) = 2 * (183m/2) * sin(71.42º/2) = 107 m

so the average velocity is

v = L / t = 107m / 7.40*60s = 0.241 m/s

Initially, let's say the velocity is along the +x axis:

Vi = π * 183m / (37.3*60s) i = 0.257 m/s i

Later, it's rotated through 71.42º, so

Vf = 0.257m/s * (cos71.42º i + sin71.42º j) = [0.0819 i + 0.244 j] m/s

ΔV = Vf - Vi = [(0.0819 - 0.257) i + 0.244 j] m/s = [-0.175 i + 0.244 j] m/s

which has magnitude

|ΔV| = √(0.175² + 0.244²) m/s = 0.300 m/s

Then the average acceleration is

a_avg = |ΔV| / t = 0.300m/s / (7.40*60s) = 6.76e-4 m/s²

The instantaneous acceleration is centripetal: a = ω²r

a = (2π rads / (37.3*60s)² * 183m/2 = 7.21e-4 m/s²

Answer:

$v = 0.24 m/s$

$a = 6.75 * 10^(-4) m/s^2$

Explanation:

Given that wheel completes one round in total time T = 37.3 min

so angular speed of the wheel is given as

$\omega = (2\pi)/(T)$

$\omega = (2\pi)/(37.3) rad/min$

now the angle turned by the wheel in time interval of t = 7.40 min

$\theta = \omega t$

$\theta = ((2\pi)/(37.3))(7.40) = 0.4\pi$

PART 1)

Now the average velocity is defined as the ratio of displacement and time

here displacement in given time interval is

$d = 2Rsin(\theta)/(2)$

R = radius = 91.5 m

$d = 183sin(0.2\pi) = 106.8 m$

Now time to turn the wheel is given as

$t = 7.40 min = 444 s$

now we have

$v = (d)/(t) = (106.8)/(444)$

$v = 0.24 m/s$

PART 2)

Now average acceleration is defined as ratio of change in velocity in given time interval

here velocity of a point on its rim is given as

$v = R\omega$

$v = (91.5)((2\pi)/(37.3* 60))$

$v = 0.257 m/s$

now change in velocity when wheel turned by the above mentioned angle is given as

$\Delta v = 2vsin(\theta)/(2)$

$\Delta v = 2(0.257)sin(0.2\pi)$

$\Delta v = 0.3 m/s$

time interval is given as

$t = 7.40 min = 444 s$

now average acceleration is given as

$a = (0.3)/(444)$

$a = 6.75 * 10^(-4) m/s^2$

The actual depth of a shallow pool 1.00 m deep is not the same as the apparent depth seen when you look straight down at the pool from above. How deep (in cm) will it appear to be

Answers

Answer:

d' = 75.1 cm

Explanation:

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

$n=(d)/(d')\n\nd'=(d)/(n)\n\nd'=(1\ m)/(1.33)\n\nd'=0.751\ m$

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

The apparent depth of a 1.00-meter-deep pool, when viewed from above, is around 75.2 centimeters. This difference is due to light refraction in water, causing optical distortion.

When observing a shallow pool of 1.00 meter depth from above, the apparent depth is altered by the phenomenon of light refraction in water. Light bends as it passes from air into water, affecting the way objects are perceived underwater.

The apparent depth is less than the actual depth due to this bending of light. To calculate the apparent depth, one can use the Snell's Law formula, which relates the angles of incidence and refraction to the refractive indices of the two media.

However, a simplified formula for the apparent depth (d') in terms of the actual depth (d) is given by d' = d/n, where 'n' is the refractive index of water (approximately 1.33). Therefore, in this case, the pool's apparent depth, when viewed from above, will be approximately 75.2 centimeters, making it shallower than it appears at first glance due to the optical effects caused by light traveling through water.

For more such questions on refraction

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Complete question below:

"What is the apparent depth, in centimeters, when looking straight down at a shallow pool that is 1.00 meter deep? Note that the apparent depth is different from the actual depth due to the refraction of light in water."

How does light move?

Answers

Answer:

Light travels as a wave. But unlike sound waves or water waves, it does not need any matter or material to carry its energy along. This means that light can travel through a vacuum—a completely airless space. It speeds through the vacuum of space at 186,400 miles (300,000 km) per second.

Explanation:

Hope this helps :))

What type of electromagnetic radiation is being shown in the picture?A. Gamma rays
B. Ultraviolet radiation
C. X-rays
D. Infrared radiation

Answers

Answer:

I think D. Infrared radiation.

Answer:

infrared radition

Explanation:

valid

A pressure antinode in a sound wave is a region of high pressure, while a pressure node is a region of low pressure.True
False

Answers

A pressure antinode in a sound wave is not a region of high pressure, while a pressure node is not a region of low pressure.

The answer is false

Final answer:

A pressure antinode in a sound wave is indeed a region of high pressure, while a pressure node is a region of low pressure. These definitions hold true for all types of waves.

Explanation:

That's true. In terms of sound waves, a pressure antinode is a region of high pressure, while a pressure node is a region of low pressure. This is true for all types of waves, not only sound waves. In essence, a wave moves through a medium (in case of a sound wave, that medium is typically air) by creating areas of high and low pressure - the high pressure areas are called antinodes, and the low pressure areas are called nodes.