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3 years ago

All but one statement applies to electromagnetic radiation

Physics

1 answer:

lina2011 [118]3 years ago

3 0

<span>D) Electromagnetic radiation travels in the form of longitudinal waves.</span>

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Please Answer the question in the picture ASAP PLEASE

attashe74 [19]

Answer:

HERE IS YOUR ANSWER

Explanation:

PLEASE MARK MY ANSWER AS BRAINLIEST IF THE ANSWERS ARE CORRECT .

Beacuse of the loose connection of the wire .

Straight

5 0

3 years ago

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a s

tatiyna

Answer:

a. λ = 647.2 nm

b. I₀ 9.36 x 10⁻⁵

Explanation:

Given:

β = 56.0 rad , θ = 3.09 ° , γ = 0.170 mm = 0.170 x 10⁻³ m

The wavelength of the radiation can be find using

β = 2 π / γ * sin θ

λ = [ 2π * γ * sin θ ] / β

λ = [ 2π * 0.107 x 10⁻³m * sin (3.09°) ] / 56.0 rad

λ = 647.14 x 10⁻⁹ m ⇒ λ = 647.2 nm

The intensity of the central maximum I₀

I = I₀ (4 / β² ) * sin ( β / 2)²

I = I₀ (4 / 56.0²) * [ sin (56.0 /2) ]²

I = I₀ 9.36 x 10⁻⁵

8 0

4 years ago

if abus travelling at 20m/s is subject to steady decceleration of 5m/s².how long will it take yo come to rest.

Likurg_2 [28]

Answer:

4 seconds

Explanation:

Given:

v₀ = 20 m/s

v = 0 m/s

a = -5 m/s²

Find: t

v = at + v₀

0 m/s = (-5 m/s²) t + 20 m/s

t = 4 s

6 0

4 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

4 years ago

A 2.55 kg block starts from rest on a rough inclined plane that makes an angle of 60 degree with the horizontal. the coefficient

lorasvet [3.4K]

Answer:

Change in mechanical energy = work done by friction

so it is equal to

$W = -8.16 J$

Explanation:

As we know that change in mechanical energy must be equal to the work done by non conservative forces only

So here when block moves down the inclined plane then the work done by friction force is given as

$W = F.d$

here we have

$F = \mu F_n$

here we know that

$F_n = mg cos\theta$

so we have

$F_n = 2.55(9.81)(cos60)$

$F_n = 12.5 N$

Now the friction force on the block is given as

$F_f = \mu F_n$

$F_f = 0.25 \times 12.5$

$F_f = 3.13 N$

now work done by the friction is given as

$W = -(3.13)(2.61)$

$W = -8.16 J$

7 0

4 years ago