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3 years ago

8

What happens to a light ray if it is incident on a reflective surface along the normal?

1 answer:

Bad White [126]3 years ago

4 0

Answer: The incident ray and the reflected ray and the normal will be parallel to each other.

Explanation:

The normal is perpendicular to the surface of the mirror or the reflective surface.

According to the law of reflection which state that:

The angle of incidence is always equal to the angle of reflection on a smooth surface.

If a light ray is incident on a reflective surface along the normal. The angle of incidence will be at 90 degrees which will be perpendicular to the surface of the mirror, the reflected ray will bounce back likewise at the same angle which will be perpendicular to the reflective surface.

Both the incident ray and the reflected ray and the normal will be parallel to each other.

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The volume flow rate of blood leaving the heart to circulate throughout the body is about 5 L/min for a person at rest. All this

skelet666 [1.2K]

Answer:

$n=2.9\times 10^9$

$A=1.88\times 10^{-8}\ m^2$

Explanation:

Given that

Q= 5 L/min

1 L = 10⁻³ m³/s

1 min = 60 s

Q=0.083 x 10⁻³ m³/s

d= 6 μm

v= 1 mm/s

So the discharge flow through one tube

q = A v

$A=\dfrac{\pi}{4}d^2$

$A=\dfrac{\pi}{4}\times (6\times 10^{-6})^2\ m^2$

A=2.8 x 10⁻¹¹ m²

v= 1 x 10⁻³ m/s

q= 2.8 x 10⁻¹⁴ m³/s

Lets take total number of tube is n

Q= n q

n=Q/q

$n=\dfrac{0.083\times 10^{-3} }{ 2.8\times 10^{-14}}$

$n=2.9\times 10^9$

Surface area A

A= π d L

$A=\pi \times 6\times 10^{-6}\times 10^{-3}\ m^2$

$A=1.88\times 10^{-8}\ m^2$

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3 years ago

Name three elements that have similar chemical properties to oxygen

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3 years ago

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Air expands adiabatically in a piston–cylinder assembly from an initial state where p1 = 100 lbf/in.2, v1 = 3.704 ft3/lb, and T1

lutik1710 [3]

Answer:

Final temperature is equal to 1291.63°R

Explanation:

given,

p₁ = 100 lb f/in², v₁ = 3.704 ft³/lb, and T₁ = 1000 °R

p₂ = 30 lb f/in² n = 1.4

Δ u = 0.171(T₂ - T₁)

we know for poly tropic process

p vⁿ = constant

p₁ v₁ⁿ = p₂ v₂ⁿ

100 × 3.704¹°⁴ = 30 × v₂¹°⁴

v₂ = 8.753 ft³/lb

work done for poly tropic process

W = $\dfrac{p_1v_1-p_2v_2}{n-1}$

= $\dfrac{100\times 3.704-30\times 8.753}{1.4-1}$

= 269.525 lbf/in².ft³

W = $\dfrac{269.525}{5.40395}$ Btu/lb

= 49.87 Btu/lb

in the piston cylinder arrangement air is expanding acrobatically

Δ q = Δu + w

Δ u = - w

0.171(T₂ - T₁) = -49.87

0.171(T₁ - T₂) = -49.87

0.171 T₂ = 0.171 × 1000 + 49.87

T₂ = 1291.63 °R

Final temperature is equal to 1291.63°R

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3 years ago