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1 year ago

What is the electric field strength between two parallel plates 5cm apart with a potential difference of 25V across them?

Physics

1 answer:

Butoxors [25]1 year ago

5 0

Use the following formula for the electric field strength between two parallel plates:

E = V/d

where,

V: potential difference = 25V

d: distance between plates = 5 cm = 0.05 m

Replace the previous values of the parameters into the formula for E:

$E=\frac{25V}{0.05m}=500\frac{V}{m}$

Hence, the electric field strength is 500V/m

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What is the energy in joules of a mole of photons associated with visible light of wavelength 486 nm?

ivann1987 [24]

Answer:

$2.46\cdot 10^5 J$

Explanation:

The energy of a single photon is given by:

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength

For the photon in this problem,

$\lambda=486 nm=4.86\cdot 10^{-7}m$

So, its energy is

$E_1=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.86\cdot 10^{-7}m}=4.09\cdot 10^{-19} J$

One mole of photons contains a number of photons equal to Avogadro number:

$N_A = 6.022\cdot 10^{23}$

So, the total energy of one mole of photons is

$E=N_A E_1 = (6.022\cdot 10^{23})(4.09\cdot 10^{-19} J)=2.46\cdot 10^5 J$

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3 years ago

A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo

CaHeK987 [17]

Answer: $\omega =93.51 rad/s$

Explanation:

Given

mass of Flywheel $m_1=1440 kg$

mass of bus $m_b=10200 kg$

radius of Flywheel $r=0.63 m$

final speed of bus $v=21 m/s$

Conserving Energy i.e.

0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus

Let $\omega$ be the angular velocity of Flywheel

$0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}$

$I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2$

$0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}$

$\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}$

$\omega =21\times 4.45=93.51 rad/s$

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2 years ago

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

levacccp [35]

Answer:

The minimum speed must the car must be 13.13 m/s.

Explanation:

The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.

We know that, mg be the weight of car and rider, which is equal to the centripetal force.

$mg = \dfrac{mv^2}{r}\\v = \sqrt{rg}\\v = \sqrt{17.6\times 9.8}\\v = 13.13\ m/s\\$

So, the minimum speed must the car must be 13.13 m/s.

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2 years ago

Why do planets rings surround them and not the sun

Shalnov [3]

Well, the rings surrounding a planet are made out of rock. A ring surrounding the sun would be impossible since the sun can reach more than 27 million degrees Fahrenheit (15 million degrees Celsius.)

Hope this helped.

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3 years ago

If a force of 26 N is applied to an object with a mass of 8 kg, the object will accelerate at m/s2

malfutka [58]

False, the numbers do not add up

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2 years ago

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