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Oksi-84 [34.3K]

3 years ago

9

A monochromatic light falls on two narrow slits that are 4.50 um apart. The third destructive fringes which are 35° apart were f

ormed at a screen 2m from the light source. The light source is 0.30 m from the slits. () Calculate Ym. (4 marks) Compute the wavelength of the light. (4 marks)

1 answer:

STatiana [176]3 years ago

3 0

Answer:

y = 1.19 m and λ = 8.6036 10⁻⁷ m

Explanation:

This is a slit interference problem, the expression for destructive interference is

d sin θ = m λ

indicate that for the angle of θ = 35º it is in the third order m = 3 and the separation of the slits is d = 4.50 10⁻⁶ m

λ = d sin θ / m

let's calculate

λ = 4.50 10⁻⁶ sin 35 /3

λ = 8.6036 10⁻⁷ m

for the separation distance from the central stripe, we use trigonometry

tan θ= y / L

y = L tan θ

the distance L is measured from the slits, it indicates that the light source is at x = 0.30 m from the slits

L = 2 -0.30

L = 1.70 m

let's calculate

y = 1.70 tan 35

y = 1.19 m

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Oil at 150 C flows slowly through a long, thin-walled pipe of 30-mm inner diameter. The pipe is suspended in a room for which th

chubhunter [2.5K]

Answer:

<em>1.01 W/m</em>

Explanation:

diameter of the pipe d = 30 mm = 0.03 m

radius of the pipe r = d/2 = 0.015 m

external air temperature Ta = 20 °C

temperature of pipe wall Tw = 150 °C

convection coefficient at outer tube surface h = 11 W/m^2-K

From the above,<em> we assumed that the pipe wall and the oil are in thermal equilibrium</em>.

area of the pipe per unit length A = $\pi r ^{2}$ = $7.069*10^{-4}$ m^2/m

convectional heat loss Q = Ah(Tw - Ta)

Q = 7.069 x 10^-4 x 11 x (150 - 20)

Q = 7.069 x 10^-4 x 11 x 130 = <em>1.01 W/m</em>

4 0

4 years ago

1. A 12 kilogram block is sitting on a platform 24 m high. How much Potential energy does

Mazyrski [523]

The amount of potential energy the block contains is 2,822.4 Joules

<u>Given the following data:</u>

Mass of block = 12 kg

Height of platform = 24 meters.

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 $m/s^2$ .

To determine the amount of potential energy the block contains:

Mathematically, potential energy (P.E) is given by the formula;

$P.E = mgh$

Where:

m is the mass of object.

g is the acceleration due to gravity.

h is the height of an object.

Substituting the parameters into the formula, we have;

$P.E = 12 \times 9.8 \times 24$

Potential energy (P.E) = 2,822.4 Joules

Read more: brainly.com/question/23153766

8 0

3 years ago

Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A p

serg [7]

Answer:

f = 735 Hz

Explanation:

given,

Person distance from speakers

r₁ = 4.1 m r₂ = 4.8 m

Path difference

d = r₂ - r₁ = 4.8 - 4.1 = 0.7 m

For destructive interference

$d = \dfrac{n\lambda}{2}$

where, n = 1, 3,5..

we know, λ = v/f

$d = \dfrac{n v}{2f}$

v is the speed of the sound = 343 m/s

f is the frequency

$f = \dfrac{n v}{2d}$

for n = 1

$f = \dfrac{343}{2\times 0.7}$

f = 245 Hz

for n = 3

$f = \dfrac{3\times 343}{2\times 0.7}$

f = 735 Hz

Hence,the second lowest frequency of the destructive interference is 735 Hz.

7 0

4 years ago

When water (H₂O) freezes into ice, some of the properties have changed. What stays the same?

erastova [34]

Answer:

1st one

identity of the H₂O

When water (H2O) freezes into ice, some of the properties have changed but the identity of the H2O is the same. Explanation; Water can exist in liquid, solid; ice and gaseous state; steam. Freezing occurs when water is continuously cool until it turns to solid ice, water freezes at zero degree Celsius

Hope This Helps

4 0

2 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

Mariana [72]

a) $F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

$R-Fsin \theta - mg = 0$ (1)

where

R is the normal reaction of the floor (upward)

$F sin \theta$ is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

$F cos \theta - \mu_k R = 0$ (2)

where

$F cos \theta$ is the horizontal component of the force F (forward)

$\mu_k R$ is the force of friction (backward)

From (1) we get

$R=Fsin \theta +mg$

And substituting into (2)

$F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

b) $\mu_s=cot(\theta)$

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

$R-Fsin \theta - mg = 0$ (1)

$F cos \theta - \mu_s R = 0$ (2)

In this second case, we want to find the critical value of $\mu_s$ such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, $F cos \theta$ .

Therefore, using the same procedure as before,

$R=Fsin \theta +mg$

$F cos \theta - \mu_s (Fsin \theta +mg) = 0$

And solving for $\mu_s$ ,

$F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}$

Now we analyze the expression that we found. We notice that if the force applied F is very large, $F sin \theta >> mg$ , therefore we can rewrite the expression as

$\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)$

So, this is the critical value of the coefficient of static friction.

8 0

3 years ago