Expression to calculate pressure

Monochromatic light with wavelength 588 nm is incident on a slit with width 0.0351 mm. The distance from the slit to a screen is

Norma-Jean [14]

Answer:

Explanation:

A. Using

Sinစ= y/ L = 0.013/2.7= 0.00481

စ=0.28°

B.here we use

Alpha= πsinစa/lambda

= π x (0.0351)sin(0.28)/588E-9m

= 9.1*10^-2rad

C.we use

I(စ)/Im= (sin alpha/alpha) ²

So

{= (sin0.091/0.091)²

= 3*10^-4

6 0

4 years ago

What naturally occurring radioisotope is used as the fuel for a nuclear reactor?

Lady_Fox [76]

The naturally occurring radioisotope which is used as the fuel for a nuclear reactor is the Uranium-235. It has been said that one kilogram of Uranium-235 can theoretically male about 20 terajoules of energy when used as a fuel for power plants.

3 0

3 years ago

How many meters per second is a wave going if it travels a distance of one mile in 0.0012 hours

PtichkaEL [24]

Answer:

Explanation I don’t know :

5 0

3 years ago

An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 2060 × 103 seconds (about

Fittoniya [83]

Answer:

Acceleration, $a=2.22\times 10^{-3}\ m/s^2$

Explanation:

It is given that,

Time period of revolution of the moon, $T=2060\times 10^3\ s$

If the distance from the center of the moon to the surface of the planet is, $h=235\times 10^6\ m$

The radius of the planet, $r=3.9\times 10^6\ m$

Let a is the moon's radial acceleration. Mathematically, it is given by :

$a=R\times \omega^2$ , R is the radius of orbit

Since, $\omega=\dfrac{2\pi}{T}$

The radius of orbit is,

$R=r+h$

$R=3.9\times 10^6\ m+235\times 10^6\ m=238900000\ m$

So, $a=\dfrac{4\pi^2 R}{T^2}$

$a=\dfrac{4\pi^2 \times 238900000}{(2060\times 10^3)^2}$

$a=2.22\times 10^{-3}\ m/s^2$

Hence, this is the required solution for the radial acceleration of the moon.

5 0

3 years ago

A 3.0N force to the right acts on a 0.5kg object at rest during a time interval of 4.0 seconds .What is the velocity of the obje

monitta

Answer:

24 m/s

Explanation:

Applying Newton's second law of motion,

F = ma................ Equation 1

Where F = force acting on the object, m = mass of the object a = acceleration of the object.

First we calculate for a.

make a the subject of the formula in equation 1 above

a = F/m............... Equation 2

Given: F = 3 N, m = 0.5 kg.

Substitute into equation 2

a = 3/0.5

a = 6 m/s²

Secondly we calculate the velocity by using,

v = u+at................... Equation 3

Where v = Final velocity of the object, u = initial velocity of the object, t = time interval

Given: u = 0 m/s (at rest), a = 6 m/s², t = 4 s

Substitute into equation 3

v = 0+6(4)

v = 24 m/s

5 0

3 years ago

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