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Otrada [13]
3 years ago
10

Monochromatic light with wavelength 588 nm is incident on a slit with width 0.0351 mm. The distance from the slit to a screen is

2.7 m. Consider a point on the screen 1.3 cm from the central maximum. Calculate (a) θ for that point, (b) α, and (c) the ratio of the intensity at that point to the intensity at the central maximum.
Physics
1 answer:
Norma-Jean [14]3 years ago
6 0

Answer:

Explanation:

A. Using

Sinစ= y/ L = 0.013/2.7= 0.00481

စ=0.28°

B.here we use

Alpha= πsinစa/lambda

= π x (0.0351)sin(0.28)/588E-9m

= 9.1*10^-2rad

C.we use

I(စ)/Im= (sin alpha/alpha) ²

So

{= (sin0.091/0.091)²

= 3*10^-4

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3 0
2 years ago
A car starts from rest and acquires a velocity of 50m/s in 3secs. Calculate i) acceleration ii) distance covered.
mafiozo [28]

Answer: 75.02 m

Explanation:

u = 0 ( starts from rest )

v = 50 m/s

t = 3 s

( i ) a = v - u / t

= 50 - 0 /3

= 16.67

( ii ) s = ut + 1/2 at²

= 0 × 3 + 1/2 × 16.67 × 3 × 3

= <u>75.02 m</u>

Hope this helps...

4 0
3 years ago
Earth's surface receives about twice as much energy from the atmosphere than from the sun as a result of ____.
Oxana [17]
I think the answer is “greenhouse effect”
5 0
3 years ago
Adding heat to a liquid causes which of the following physical changes?
SashulF [63]

In almost every case in nature, adding heat to a liquid
causes the density of the liquid to decrease.  That is,
when the liquid gets warmer, it expands and occupies
more space.

The one big exception to this rule is water ! 

Starting with a block of ice at zero°C (32°F), as the ice melts,
becomes water at zero°C, and all the way to 4°C (about 39°F),
its density increases all the way.   That is, it shrinks and occupies
less
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This sounds like an interesting but insignificant quirk ... until
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8 0
3 years ago
A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t
Mumz [18]

Answer:

v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t

Second stone:

y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

6 0
3 years ago
Read 2 more answers
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