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2 years ago

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Assume your eye has an aperture diameter of 3.00 mm at night when bright headlights are pointed at it. 1) At what distance can y

ou see two headlights separated by 1.80 m as distinct? Assume a wavelength of 550 nm, near the middle of the visible spectrum. (Express your answer to two significant figures.)

2 answers:

Sauron [17]2 years ago

5 0

Answer:

8.1 x 10³ m

Explanation:

D = diameter of the aperture =3 mm = 0.003 m

x = distance between the two headlights = 1.80 m

λ = wavelength of the visible spectrum = 550 x 10⁻⁹ m

d = distance of the headlights from eyes = ?

Using Reyleigh's criterion,

$\frac{x}{d}=\frac{1.22\lambda }{D}$

$\frac{1.80}{d}=\frac{1.22 (550\times 10^{-9}) }{0.003}$

d = 8047.7 m

d = 8.1 x 10³ m

krek1111 [17]2 years ago

3 0

Answer:

The distance is 6259.31 meters.

Explanation:

We shall use the Reyligh criterion to solve the problem

For diffraction due to circular aperture we have

Assuming that human eye is circular we have

$\frac{x}{D}=\frac{1.22\lambda }{d}$

$\therefore D=\frac{xd}{1.22\lambda }$

Applying the given values we have

$D=\frac{1.40\times 3\times 10^{-3}}{1.22\times 550\times 10^{-9}}\\\\\therefore D=6259.31m\\D=6.26km$

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Give two ways of reversing the direction of the forces on the coil in the electric motor?

nikklg [1K]

Answer:

Interchanging the poles of the magnet

Reversing the direction of the applied current

Explanation:

The working of the electric motor is associated with Fleming's left-hand rule.
It states that if a current-carrying conductor is placed inside a magnetic field, it experiences a force in the direction perpendicular to the direction of the electric current and magnetic field.
These three physical quantities are placed in a mutually perpendicular direction.
So, in order to reverse the direction of force, you have to either change the direction of the current or magnetic field.

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2 years ago

Steam at 700 bar and 600 oC is withdrawn from a steam line and adiabatically expanded to 10 bar at a rate of 2 kg/min. What is t

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Answer:

Final temperature of the steam = 304.29 K = 31.14°C

Rate Entropy generation of the process should be 0 because no heat is transferred into or out of the system. And ΔS = Q/T = 0 J/K.

But 0.01 J/k.s was obtained though, which is approximately 0.

Explanation:

For an adiabatic system, the Pressure and temperature are related thus

P¹⁻ʸ Tʸ = constant

where γ = ratio of specific heats. For steam, γ = 1.33

P₁¹⁻ʸ T₁ʸ = P₂¹⁻ʸ T₂ʸ

P₁ = 700 bar

P₂ = 10 bar

T₁ = 600°C = 873.15 K

T₂ = ?

(700⁻⁰•³³)(873.15¹•³³) = (10⁻⁰•³³)(T₂¹•³³)

T₂ = 304.29 K = 31.14°C

b) Rate Entropy generation of the process should be 0 because no heat is transferred into or out of the system. And ΔS = Q/T

To prove this

Entropy of the process

dQ - dW = dU

dQ = dU + dW

dU = mCv dT

dW = PdV

dQ = TdS

TdS = mCv dT + PdV

dS = (mCv dT/T) + (PdV/T) =

PV = mRT; P/T = mR/V

dS = (mCv dT/T) + (mRdV/V)

On integrating, we obtain

ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)

To obtain V₁ and V₂, we use PV = mRT

V/m = specific volume

Pv = RT

R for steam = 461.52 J/kg.K

For V₁

P = 700 bar = 700 × 10⁵ Pa, T = 873.15 K

v₁ = (461.52 × 873.15)/(700 × 10⁵) = 0.00570 m³/kg

For V₂

P = 10 bar = 10 × 10⁵ Pa, T = 304.29 K

v₂ = (461.52 × 304.29)/(10 × 10⁵) = 0.143 m³/kg

ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)

m = 2 kg/min = 0.0333 kg/s, Cv = 1410.8 J/kg.K

ΔS = [0.03333 × 1410.8 × In (304.29/873.15)] + (0.03333 × 461.52 × In (0.143/0.00570)

ΔS = - 49.56 + 49.57 = 0.01 J/K.s ≈ 0 J/K.s

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At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is

Nastasia [14]

At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
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v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²

Point 1:
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v₁ = 150 ft/s

Point 2 (stagnation):
At the stagnation point, the velocity is zero.

The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
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borishaifa [10]

Answer: k= $\frac{5mv^{2} }{2}$

Explanation:

Recall that the formula for kinetic energy is given below as

k = $\frac{mv^{2} }{2}$

where k=kinetic energy (joules), m= mass of object (kg), v= velocity of object m/s)

For cart A

$m_{a}$ = mass of cart A

$v_{a}$ = v = velocity of cart A

$K.E_{a}$ = kinetic energy of cart A

hence, $K.E_{a}$ = $\frac{m_{a}v^{2} }{2}$

For cart B

$m_{b}$ = mass of cart B

$v_{b}$ = 2v = velocity of cart B

$K.E_{b}$ = kinetic energy of cart B

hence, $K.E_{b}$ = $\frac{m_{b}(2v^{2}) }{2}$ = 2 $m_{b} v^{2}$

from the question, both cart are identical which implies they have the same mass i.e $m_{a}$ = $m_{b}$ = m which implies that

$K.E_{a}= \frac{mv^{2} }{2}$ and $K.E_{b} =2mv^{2}$

The total kinetic energy K is the sum of cart A and cart B kinetic energy

$K=K.E_{a} + K.E_{b}$

$K=\frac{mv^{2} }{2} + 2mv^{2}$

hence

$K=\frac{5mv^{2} }{2}$

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<h2>Answer: at an angle $\theta$ below the inclined plane. </h2>

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$W=m.g$

This force is directed vertically at an angle $\theta$ below the inclined plane, this means it has an X-component and a Y-component:

$W=W_{X}+W_{Y}$

$W_{X}=m.g.cos(\theta)$

$W_{Y}=m.g.sin(\theta)$

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