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sweet [91]
3 years ago
15

Light with a wavelength of λ = 689 nm. is incident on a single slit of width w = 3.75 micrometers. A screen is located L = 0.65

m behind the slit and an interference pattern has formed on it. What is the distance between the central bright spot and the first dark fringe, D, in meters?
Physics
1 answer:
rodikova [14]3 years ago
8 0

Answer:

119.42 x 10⁻³ m

Explanation:

The distance of first dark fringe from central bright spot will be equal to the width of fringe .

Fringe width = λL /w

where λ is wavelength of light, w is slit width , L is distance of screen .

So required distance

= \frac{689\times10^{-9}\times0.65}{3.75\times10^{-6}}

= 119.42 x 10⁻³ m

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A cell phone charger operates at 5 volts and can deliver a maximum current of 1.0 Amp. If your cell phone has an energy storage
madam [21]

Answer:

<em>it</em><em> </em><em>takes</em><em> </em><em>2</em><em>.</em><em>1</em><em>6</em><em> </em><em>hrs</em><em> </em><em>to</em><em> </em><em>fully</em><em> </em><em>charge</em><em> </em><em>the</em><em> </em><em>phone</em>

Explanation:

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I = 1A

E = 38.9 Kj = 38900 j

t = ?

E = IVt

t = E/IV

t = 38900 ÷ (1 × 5)

t = 38900 ÷ 5

t = 7780 seconds

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7 0
3 years ago
How to measure the density of a white dwarf if its diameter is 1000 km and weighs 10³⁰ kg?
Afina-wow [57]

Well, first of all, 10³⁰ kg is its mass, not its weight.

-- Density of anything is (its mass) divided by (its volume).

-- You know the white dwarf's mass.

-- The volume of any sphere is  (4/3) (pi) (radius)³

-- The radius is  1/2  of the diameter.

--  Now you have the ball.  Run with it.

I did it quickly and got  (1.91 x 10¹²) kilogram/meter³ .
That may be wrong.  You have to check it.  And even
if it's correct, you probably want to express density in
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6 0
3 years ago
The pressure of the earth's atmosphere at sea level is 14.7 lb/in2. What is the pressure when expressed in g/m2? (2.54 cm = 1 in
miss Akunina [59]
<h2>Answer:</h2>

14.7 lb / in² = 10333018.166 g / m²

<h2>Explanation:</h2>

Given from the question;

pressure = 14.7lb/in²

<u>To convert from lb / in² to g / m²</u>

Note;

2.54 cm = 1 in       ------------(i)

But;

100 cm = 1m

=> 2.54cm = 2.54 cm x 1 m / 100 cm = 0.0254m

=>2.54cm = 0.0254m

Substitute 2.54cm = 0.0254m into equation(i)

=> 0.0254m = 1 in

<em>Note also;</em>

2.205 lb = 1 kg            ------------------(ii)

But;

1kg = 1000g

Substitute 1kg = 1000g into equation (ii)

=> 2.205 lb = 1000g   ---------------------(iii)

<em>From equation(iii);</em>

if, 2.205 lb = 1000g

then, 1 lb = 1 lb x 1000 g / 2.205 lb = 453.5g

=> 1 lb = 453.5 g

<em>Now let's convert 14.7 lb/in² to g / m²;</em>

=> 14.7 lb / in² could be written as 14.7 x 1 lb / (1 in x 1 in)

i.e

=> 14.7 lb / in² = 14.7 x 1 lb / (1 in x 1 in)     ------------------(iv)

<em>Substitute the values of 1 lb = 453.5g and 1 in = 0.0254m into equation(iv)</em>

=> 14.7 lb / in² = 14.7 x 453.5g / (0.0254m x 0.0254m)

=> 14.7 lb / in² = 6666.45 g / (0.00064516m²)

=> 14.7 lb / in² = 10333018.166 g / m²

<em>Therefore, 14.7 lb / in² = 10333018.166 g / m²</em>

5 0
3 years ago
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Vilka [71]

Answer:

I = 5[amp]

Explanation:

Electrical power is defined as the product of voltage by current.

P=V*I

where:

P = power = 1150 [W]

V = voltage = 230 [V]

I = current [amp]

Now replacing:

1150=230*I\\I=1150/230\\I=5[amp]

A 15 [amp] fuse must be used. Always the fuse must be larger than the operating current, to protect the equipment from very high currents. above 15 [amp]

4 0
3 years ago
11 Design Imagine that a scientistdiscovered a way to make africtionless surface. What wouldbe some useful applications forthis
m_a_m_a [10]

Answer:

You could move something across the Earth with a little push. It would make fuel really efficient on those pathways. You could make a floor that is impossible to walk on. Everybody would just fall without traction.

Explanation:

3 0
3 years ago
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