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Murljashka [212]
3 years ago
13

Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the

third bright fringe occurs at ± 3.10 cm on either side of the central bright fringe. How far apart are the two slits?
Physics
1 answer:
IgorC [24]3 years ago
7 0

Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency f= 6.37\times10^{14}\ Hz

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{c}{f}

where, c = speed of light

f = frequency

Put the value into the formula

\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}

\lambda=471\ nm

We need to calculate the distance between the two slits

m\times \lambda=d\sin\theta

d =\dfrac{m\times\lambda}{\sin\theta}

Where, m = number of fringe

d = distance between the two slits

Here, \sin\theta =\dfrac{y}{l}

Put the value into the formula

d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}

d=40.11\times10^{-6}\ m

d = 40.11\ \mu m

Hence, The distance between the two slits is 40.11 μm.

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At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
lbvjy [14]

Answer:

F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

|F_T|=2\sqrt{34}k\frac{Q^2}{L}

\theta=tan^{-1}(\frac{5}{3})=59.03\°

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

and the magnitude is:

|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}

the direction is:

\theta=tan^{-1}(\frac{5}{3})=59.03\°

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you have to draw diagram

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2 years ago
To win the game, a placekicker must kick a football from a point 44 m (48.1184 yd) from the goal, and the ball must clear the cr
Ganezh [65]

Answer:

The distance by the ball clear the crossbar is 1.15 m

Explanation:

Given that,

Distance = 44 m

Speed = 24 m/s

Angle = 31°

Height = 3.05 m

We need to calculate the horizontal velocity

Using formula of horizontal velocity

u_{x}=u\cos\theta

Put the value into the formula

u_{x}=24\cos(31)

u_{x}=20.5\ m/s

We need to calculate the vertical velocity

Using formula of vertical velocity

u_{y}=u\sin\theta

Put the value into the formula

u_{y}=24\sin(31)

u_{y}=12.3\ m/s

We need to calculate the time

Using formula of time

t=\dfrac{d}{u_{x}}

Put the value into the formula

t=\dfrac{44}{20.5}

t=2.1\ sec

We need to calculate the vertical height

Using equation of motion

h=u_{y}t+\dfrac{1}{2}at^2

Put the value into the formula

h=12.3\times2.1-\dfrac{1}{2}\times9.8\times(2.1)^2

h=4.2\ m

We need to calculate the distance by the ball clear the crossbar

Using formula for vertical distance

d=h-3.05

Put the value of h

d=4.2-3.05

d=1.15\ m

Hence, The distance by the ball clear the crossbar is 1.15 m

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Toby's Trucking Company determined that on an annual basis, the distance traveled per truck is normally distributed, with a mean
Reptile [31]

Answer: 56994 miles

Explanation:

Given : Toby's Trucking Company determined that on an annual basis, the distance traveled per truck is normally distributed with

\mu=50,000\text{ miles}

Standard deviation : \sigma=12,000\text{ miles}

Let a be the distance traveled by at least 72% of the trucks.

Let X be the random variable that represents the distance traveled by a truck

Then P(x\geq a)=0.72

The critical value corresponds to p-value 0.72 :z=0.5828415

Also, z=\dfrac{x-\mu}{\sigma}

\Rightarrow\ 0.5828415=\dfrac{a-50000}{12000}\\\\\Rightarrow\ a=12000(0.5828415)+50000=56994.098\approx56994

Hence, 56994 miles will be traveled by at least (equal to and more than) 72% of the trucks .

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What is the gravitational relationship between two objects?
Inessa [10]
I think the right answer would be objects pull because gravitational pull is when an object with more mass than an other object would pull the small mass object
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3 years ago
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