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Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits

Answers

Answer 1

Answer:

Answer:

The maximum number of bright spot is $n_(max) =5001$

Explanation:

From the question we are told that

The slit distance is $d = 1 \ mm = 0.001 \ m$

The wavelength is $\lambda = 400 \ nm = 400*10^(-9 ) \ m$

Generally the condition for interference is

$n * \lambda = d * sin \theta$

Where n is the number of fringe(bright spots) for the number of bright spots to be maximum $\theta = 90$

=> $sin( 90 )= 1$

$n = (d )/(\lambda )$

substituting values

$n = ( 1 *10^(-3) )/( 400 *10^(-9) )$

$n = 2500$

given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as

$n_(max) = 2 * n + 1$

The 1 here represented the central bright spot

$n_(max) = 2 * 2500 + 1$

$n_(max) =5001$

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A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.777 m and 2.67 kg, respectively. When the propellor rotates at 573 rpm (revolutions per minute), what is its rotational kinetic energy?

Answers

The formula for the rotational kinetic energy is

$KE_(rot) = (1)/(2)(number \ of\ propellers)( I)( omega)^(2)$

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

$I=m L^(2)=(2.67 \ kg) (0.777 \ m)^(2) =2.07459 \ kgm^(2)$

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

$KE_(rot) =( (1)/(2) )(5)(2.07459 \ kgm^(2)) (60\ rad/s)^(2)$

$KE_(rot) =18,671.31 \ J$

Answer:

4833J

Explanation:

Length=0.777

mass=2.67

# rods= 5

ω=573 rpm--> $573*2\pi *(1)/(60) =60$ rad/s

I= $(1)/(3) mL^2=(1)/(3) (2.67kg)(0.777m)^2=0.537$ kgm^2

K=1/2(number of rods)(I)(ω)= $(1)/(2) *(5)(0.537)(60)^2=4833$ J

I know it's very late, but hope this helps anyone else trying to find the answer.

A runner first runs a displacement A of 3.20 km due south, and then a second displacement B that points due east. (a) The magnitude of the resultant displacement A + B is 5.38 km. What is the magnitude (in m) of B?

Answers

Answer: 4,438.96m

Explanation:

(kindly find attachment below)

From the attachment below, it can be seen that the resultant displacement and the other 2 displacements form a right angle triangle, with A+B as the hypotenus, 3.2km as the opposite and the displacement B as the adjacent.

By using phythagoras theorem

H² = O² + A²

(5.38)² = (3.20)² + B²

28.944 = 10.24 + B²

B² = 28.944 - 10.24

B² = 18.7044

B = √18.7044

B = 4.439km to meter is 4.439 * 1000 = 4,438. 96m

Answer:

B = 4325 m

Explanation:

Resolving the displacement into x and y components.

Let north = positive y component

East = positive x component

So,

Rx = B

Ry = -3.20 km

Magnitude of the resultant displacement is

R = √(B^2 + (-3.20)^2)

R is given as R = 5.38 km

Making B the subject of formula;

B = √(R^2 - (-3.20)^2)

B = √(5.38^2 - (-3.20)^2)

B = 4.325 km

B = 4325 m

Which two types of simple machines can be found in a bicycle?

Answers

Answer:

The correct answer is A) lever and wheel and axle

Explanation:

I took the quiz

hope this helps :)

As an object in motion becomes heavier, its kinetic energy _____. A. increases exponentially B. decreases exponentially C. increases proportionally D. decreases proportionally

Answers

the answer is c. as an object is in motion speeds up or “becomes heavier”, it’s kinetic energy increases proportionally: double the velocity and you quadruple the kinetic energy. this is why a tiny bullet traveling at high speed does so much more damage than a huge truck bumping into something at 1 mph. so the answer is c

Answer:

Explanation:

*Which of the following cannot be an example of projectile motion
A. A football flying through the air
B. An apple falling from a tree
C. A pencil rolling on the ground
D.A rocket dropping from its maximum height

Answers

A. football flying through the air

A cause it flying through the also a projectile is a object flying in the air like a arrow for example/also can I get Brainly

An AC voltage source has an output of ∆V = 160.0 sin(495t) Volts. Calculate the RMS voltage. Tries 0/20 What is the frequency of the source? Tries 0/20 Calculate the voltage at time t = 1/106 s. Tries 0/20 Calculate the maximum current in the circuit when the generator is connected to an R = 53.8 Ω resistor.

Answers

Answer:

RMS voltage is 113.1370 V

frequency is 780.685 Hz

voltage is −158.66942 V

maximum current is 2.9739 A

Explanation:

Given data

∆V = 160.0 sin(495t) Volts

so Vmax = 160

and angular frequency = 495

time t = 1/106 s

resistor R = 53.8 Ω

to find out

RMS voltage and frequency of the source and voltage and maximum current

solution

we know voltage equation = Vmax sin ωt

here Vmax is 160 as given equation in question

so RMS will be Vmax / √2

RMS voltage = 160/ √2

RMS voltage is 113.1370 V

and frequency = angular frequency / 2π

so frequency = 497 / 2π

frequency is 780.685 Hz

voltage at time (1/106) s

V(t) = 160.0 sin(495/ 108)

voltage = −158.66942 V

so current from ohm law at resistor R 53.8 Ω

maximum current = voltage max / resistor

maximum current = 160 / 53.8

maximum current = 2.9739 A

Final answer:

The root-mean-square voltage of the AC source is 113.14 V, its frequency is 78.75 Hz, and the voltage at time t = 1/106 s is approximately 150.4 V. The current at this peak voltage, when connected to a resistor of 53.8 Ω, is approximately 2.97 A.

Explanation:

The output of an AC voltage source can be represented by the equation V = V₀ sin ωt, where V₀ is the peak voltage, ω is the angular frequency, and t is the time. In this case, V₀ = 160 V and ω = 495 (1/s). The root-mean-square voltage (Vrms), which is commonly used to express AC voltage, can be calculated from the peak voltage using the formula Vrms = V₀/√2 which gives approximately 113.14 V.

The frequency of the source is related to the angular frequency by the equation f = ω/2π, which gives a frequency of approximately 78.75 Hz. To find the voltage at a specific time t = 1/106 s, we substitute these values into the initial equation resulting in V = V₀ sin ωt = approximately 150.4 V.

Finally, the resistance R = 53.8 Ω allows us to calculate the maximum current in the circuit given by I = V/R. The maximum current occurs at the peak voltage, so I(max) = V₀/R = approximately 2.97 A.

Learn more about AC Voltage Source here:

brainly.com/question/32877157

#SPJ3

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