PHYSICS COLLEGE

An element emits light at two nearly equal wavelengths, 577 nm and 579 nm If the light is normally incident on a diffraction grating with 2000 lines/cm., what is the distance between the 3rd order fringes of the two wavelengths on a screen 1 m from the grating?

Answers

Answer 1

Answer:

Answer:

Explanation:

d = width of slit = 1 / 2000 cm =5 x 10⁻⁶ m

Distance of screen D = 1 m.

wave length λ₁ and λ₂ are 577 x 10⁻⁹ and 579 x 10⁻⁹ m.respectively.

distance of third order bright fringe = 3.5 λ D/d

for 577 nm , this distance = 3.5 x 577 x 10⁻⁹ x 1 /5 x 10⁻⁶

= .403 m = 40.3 cm

For 579 nm , this distance = 3.5 x 579 x 1 / 5 x 10⁻⁶

= 40.5 cm

Distance between these two = 0.2 cm.

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A rabbit is moving in the negative x-direction at 1.10 m/s when it spots a predator and accelerates to a velocity of 13.9 m/s along the negative y-axis, all in 1.20 s. Determine the x-component and the y-component of the rabbit's acceleration. (Enter your answers in m/s2. Indicate the direction with the signs of your answers.) HINT

Answers

Answer:

Explanation:

Initial velocity u = -1.1 i m /s ( along - ve direction )

final velocity = - 13.9 j

change in velocity = -13.9 j + 1.1 i

rate of change of velocity

acceleration

= (-13.9 j + 1.1 i) / 1.2

= -11.58 j + 0.916 i

x component

= 0.916 m /s²

y component

= - 11.58 m / s²

You have a grindstone (a disk) that is 105.00 kg, has a 0.297-m radius, and is turning at 71.150 rpm, and you press a steel axe against it with a radial force of 46.650 N. Assuming the kinetic coefficient of friction between steel and stone is 0.451. How many turns will the stone make before coming to rest?

Answers

Answer:

3.27 turns

Explanation:

To find how many turns (θ) will the stone make before coming to rest we will use the following equation:

$\omega_(f)^(2) = \omega_(0)^(2) + 2\alpha*\theta$

Where:

$\omega_(f)$ : is the final angular velocity = 0

$\omega_(0)$ : is the initial angular velocity = 71.150 rpm

α: is the angular acceleration

First, we need to calculate the angular acceleration (α). To do that, we can use the following equation:

$\alpha = (\tau)/(I)$

Where:

I: is the moment of inertia for the disk

τ: is the torque

The moment of inertia is:

$I = (mr^(2))/(2)$

Where:

m: is the mass of the disk = 105.00 kg

r: is the radius of the disk = 0.297 m

$I = (105.00 kg*(0.297 m)^(2))/(2) = 4.63 kg*m^(2)$

Now, the torque is equal to:

$\tau = -F x r = -\mu*F*r$

Where:

F: is the applied force = 46.650 N

μ: is the kinetic coefficient of friction = 0.451

$\tau = -\mu*F*r = -0.451*46.650 N*0.297 m = -6.25 N*m$

The minus sign is because the friction force is acting opposite to motion of grindstone.

Having the moment of inertia and the torque, we can find the angular acceleration:

$\alpha = (-6.25 N*m)/(4.63 kg*m^(2)) = -1.35 rad/s^(2)$

Finally, we can find the number of turns that the stone will make before coming to rest:

$0 = \omega_(0)^(2) + 2\alpha*\theta$

$\theta = -((\omega_(0))^(2))/(2\alpha) = -((71.150 (rev)/(min))^(2))/(2*(-1.35 (rad)/(s^(2)))*(1 rev)/(2\pi rad)*((60 s)^(2))/((1 min)^(2))) = 3.27 rev = 3.27 turns$

I hope it helps you!

Unpolarized light is passed through an optical filter that is oriented in the vertical direction. 1) If the incident intensity of the light is 90 W/m2, what is the intensity of the light that emerges from the filter? (Express your answer to two significant figures.)

Answers

Answer:

45 W/m^2

Explanation:

Intensity of light, Io = 90 W/m^2

According to the law of Malus

$I=I_(0)Cos^(2)\theta$

The average value of Cos^θ is half

So, I = Io/2

I = 90 /2

I = 45 W/m^2

Final answer:

Unpolarized light, when passed through a polarizer, reduces its intensity by half. So, the intensity if the light that emerges from a vertical filter will be 45 W/m².

Explanation:

Given that the incident intensity of the unpolarized light is 90 W/m², when passed through a vertically oriented optical filter, the emerging light will be polarized and will have its intensity halved as it's the property of a polarizing filter to decrease the intensity of unpolarized light by a factor of 2. The formula used in this process is I = Io cos² θ. In the case of unpolarized light passing through a single polarizer, θ is 0. So, the formula simplifies to I = Io/2.

Therefore, the intensity of the light that emerges from the vertically oriented optical filter is: I = 90 W/m² / 2 = 45 W/m².

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We start with 5.00 moles of an ideal monatomic gas with an initial temperature of 129 ∘C. The gas expands and, in the process, absorbs an amount of heat equal to 1180 J and does an amount of work equal to 2160 J . What is the final temperature Tfinal of the gas? Use R = 8.3145 J/(mol⋅K) for the ideal gas constant.

Answers

The final temperature of an ideal monatomic gas with an initial temperature of 128°C. is 114.53°C.

From the first law of thermodynamics,

ΔU=Q - W

Where,

ΔU - change in internal energy

Q - energy absorbed

W - work

So,

ΔU = 1180 J - 2020 J

ΔU = -840 J

From ideal gas law

$\bold {\Delta U = \frac 32n R (T_2- T_1)}}\n\n\bold {T_ 2 = \frac {2\Delta U}{3nR} +T_1}$

Where, T2 is the final temperature,

n- moles of gas

R - gas constant

T1 - initial temperature,

Put the values in the equation

$\bold {T_ 2 = \frac {2* -840\ J )}{3* 5 * 8.314\ J/mol.K} + 128^oC}\n\n\bold {T_2 = 114.53 ^oC}$

Therefore, the final temperature of an ideal monatomic gas with an initial temperature of 128°C. is 114.53°C.

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The solution is in the attachment

How does the mass of an object affect its motion through the air?

Answers

The motion of an object through the air does not affect by its mass. The rate of fall of objects does not depend upon the mass.

What are free fall and air resistance?

Free fall is a motion of a body in which gravity is the only force acting upon it. An object moving upwards might not be considered to be falling. But if the object is under the effect of the force of gravity, it is said to be in free fall.

Free fall is a type of motion in which the force acting upon an object is only gravity. Objects are not encountering a significant force of airresistance as they are only falling under the sole influence of gravity. All objects under such conditions will fall with the same rate of acceleration, regardless of their masses.

As an object falls through the air, have gone through some degree of air resistance. Air resistance is the collisions of the object's leading surface with molecules present in the air. The two most common factors that have a direct effect on the amount of air resistance are the cross-sectional area of the object and the speed of the object.

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Mass does not affect the speed of falling objects assuming there is only gravity acting on it. For example Both Bullets will strike the ground at the same time. The horizontal force applied does not affect the downward motion of the bullets — only gravity and friction (air resistance), which is the same for both bullets
(Sorry if u don’t get what I mean)

A 75 kg man starts climbing a ladder that leans against a wall. If the weight of the ladder is negligible, determine how far up the ladder the man can climb before the ladders starts to slip. The coefficient of friction on both surfaces is μS=0.25

Answers

The man can climb $\bold { X (max) = 0.25* L* tan \alpha }$ , before the ladders starts to slip.

A - point at the top of the ladder

B - point at the bottom of the ladder

C - point where the man is positioned in the ladder

L- the length of the ladder

α - angle between ladder and ground

x - distance between C and B

The forces act on the ladder,

Horizontal reaction force (T) of the wall against the ladder

Vertical (upward) reaction force (R) of ground against the ladder.

Frictionalhorizontal ( to the left ) force (F)

Vertical( downwards) of the man,

mg = 75 Kg x 9.8 m/s² = 735 N

in static conditions,

∑Fx = T - F = 0 Since, T = F

∑Fy = mg - R = 0 Since, 735 - R = 0, R = 735

∑ Torques(b) = 0

In point B the torque produced by forces R and F is Zero

Then:

∑Torques(b) = 0

And the arm lever for each force,

mg = 735

Since, ∑Torques(b) = 0

$\bold {735* x* cos\alpha = F* L* sin\alpha }$ Since,T = F

$\bold {F = \frac {735* x* cos\apha }{L* sin\alpha }}$ $\bold { \frac {cos \alpha } { sin\alpha }= cot\alpha =\frac 1{tan\alpha}}$

$\bold {F = \frac {735* x* cos\apha }{L }}$

$\bold {F = 735* x* tan\alpha }}$

F < μR the ladder will starts slipping over the ground

μ(s) = 0.25

$\bold { X (max) = 0.25* L* tan \alpha }$

Therefore, the man can climb $\bold { X (max) = 0.25* L* tan \alpha }$ , before the ladders starts to slip. \

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Answer:

x (max) = 0,25*L*tanα

Explanation:

Letá call

A: point at the top of the ladder

B: the point at the bottom of the ladder

C: point where the man is up the ladder

L the length of the ladder

α angle between ladder and ground

"x" distance between C and B

Description

The following forces act on the ladder

Point A: horizontal (to the right) reaction (T) of the wall against the ladder

Point B : Vertical (upwards) reaction (R) of ground against the ladder

frictional horizontal ( to the left ) force (F)

Point C : Weight (vertical downwards)) of the man mg

mg = 75 Kg * 9,8 m/s² mg = 735 [N]

Then in static conditions:

∑Fx = T - F = 0 ⇒ T = F

∑Fy = mg - R = 0 ⇒ 735 - R = 0 ⇒ R = 735

∑Torques(b) = 0

Note: In point B the torque produced by forces R and F are equal to 0

Then:

∑Torques(b) = 0

And the arm lever for each force is:

mg = 735

d₁ for mg and d₂ for T

cos α = d₁/x then d₁ = x*cosα

sin α = d₂ / L then d₂ = L*sinα

Then:

∑Torques(b) = 0 ⇒ 735*x*cosα - T*L*sinα = 0

735*x*cosα = T*L*sinα

T = F then 735*x*cosα = F*L*sinα

F = (735)*x*cosα/L*sinα cos α / sinα = cotgα = 1/tanα

F = (735)*x*cotanα/L or F = (735)*x/L*tanα

When F < μ* R the ladder will stars slippering over the ground

μ(s) = 0,25 0,25*R = 735*x/L*tanα

x = 0,25*R*tanα*L/735

But R = mg = 735 then

0,25*L*tanα = x

Then x (max) = 0,25*L*tanα

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