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Alex17521 [72]
3 years ago
14

A parallel beam of light from a laser with a wavelength 450 nm, falls on a grating whose slits are 1.28 x 10^-4 cm apart. How fa

r (in cm) is the 1t order bright spot from the center of the pattern on a screen 3.4 m away?
Physics
1 answer:
Katena32 [7]3 years ago
7 0

Answer:

Y_1 = 1.195 m = 119.5 cm

Explanation:

For nth order bright fringe, the distance from the center of the pattern is given by :-

Y_n = \frac{n\lambda D}{d}.

Where,

n = order of nth  bright fringe. = 1

\lambda = wavelength of  given light  =  450 nm = 450*10-{9} m

D = distance between the screen and slits. =  3.4 m

and d = separation between the slits. = 1.28 *10^{-4} cm =1.28 *10^{-6} m

Y_1 = \frac{(1 *450 *10{-9} *3.4)}{(1.28*10{-6})}

Y_1 = 1.195 m = 119.5 cm

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<span>The speed of sound is dependent on how close together the molecules of the transmitting medium is.</span>
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3 years ago
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If you swim with the current in a river, your speed is increased by the speed of the water; if you swim against the current, you
Blababa [14]

Answer:

11.23%

Explanation:

Lets take

Speed of man in still water =u= 1.73 m/s

Speed of flow of water = v=0.52 m/s

When swims in downward direction then speed of man = u + v

When swims in upward direction then speed of man = u - v

Lets time taken by man when he swims in downward direction is t_1 and when he swims in downward direction is t_2

Lets distance is d and it will be remain constant in both the case

d=(u+v)t_1

d=(u-v)t_2

(1.73+0.52)t_1=(1.73-0.52)t_2

t_2=1.85t_1

Time taken in still water

2 d= t x 1.73

t=1.15 x d sec

t_1=0.44d\ sec

t_2=0.82d\ sec

total time in current = 0.82 +0.44 d=1.26 d sec

So the percentage time

percentage\ time =\dfrac{1.28-1.15}{1.15}

 Percentage time =11.32%

So it will take 11.32% more time as compare to still current.

5 0
3 years ago
A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi
Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L)  ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L

Finally, using the Parallel Axis Theorem, we calculate I_B:

I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4}  ML^{2} =\frac{1}{3} ML^{2}

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________ occurs when an electric current is made by a changing magnetic field
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Two objects, one with a mass of m and one with a mass of 4m are attracted to each other by a gravitational force. If the force o
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By 2nd Newton's Law.

F =m*g

For 1st body

Let it's force be x

thus,

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For mass 2,

F = 4m*g

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For the first body's force interms of second body substitute eqn 2 into 1

x = m*(F/4m)

x = <u>m</u><u>*</u><u>F</u>

4m

x = F/4

Therefore, the force on mass m interms of F is

F/4.

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