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iogann1982 [59]
3 years ago
11

An inventor claims to have developed a food freezer that, in steady-state conditions, requires a power input of 0.25 kW to extra

ct energy by heat transfer at a rate of 3050 J/s from the freezer contents, which are at a temperature of 270 K. Determine if this claim is real considering an ambient temperature of 293 K. (a) Can the freezer operation in such conditions
Physics
1 answer:
WARRIOR [948]3 years ago
4 0

Answer:

The inventors  claim is not real

a)  No the the freezer cannot operate in such conditions

Explanation:

From the question we are told that

     The  power input is  P_i  = 0.25 kW  =  0.25 *10^{3} \ W

      The  rate of heat transfer J  =  3050 J/s

       The temperature of the freezer content is T = 270 \ K

       The  ambient temperature is  T_a  =  293 \ K

Generally the coefficient of performance of a refrigerator at idea conditions is mathematically represented as

      COP  =  \frac{T }{Ta - T}

substituting values

     COP  =  \frac{270 }{293 - 270}

     COP  =11.7

Generally the coefficient of performance of a refrigerator at real conditions is mathematically represented as

       COP  =  \frac{J}{P_i}

substituting values

       COP  =  \frac{3050}{0.25 *10^{3}}

       COP  = 12.2

Now given that the COP  of an ideal refrigerator is  less that that of a real refrigerator then the claims of the inventor is rejected

This is because the there are loss in the real refrigerator cycle that are suppose to reduce the COP compared to an ideal refrigerator cycle where there no loss that will reduce the COP

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Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shine
TiliK225 [7]

Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because,  the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by tanθ = y/D

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ =  λD/d and y₋₁ =  -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ =  3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ =  λD/d and y₂ =  2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

7 0
3 years ago
INTEGRATED SCIENCE-8th grade science <br> Please help report cards are due for me at dec.18
serious [3.7K]

Answer:

Our drinking water comes from lakes, rivers and groundwater. For most Americans, the water then flows from intake points to a treatment plant, a storage tank, and then to our houses through various pipe systems. A typical water treatment process.

Explanation:

7 0
3 years ago
Assume that the driver begins to brake the car when the distance to the wall is d=107m, and take the car's mass as m-1400kg, its
Evgen [1.6K]

Answer:

Explanation:

a ) Let let the frictional force needed be F

Work done by frictional force = kinetic energy of car

F x 107 = 1/2 x 1400 x 35²

F = 8014 N

b )

maximum possible static friction

= μ mg

where μ is coefficient of static friction

= .5 x 1400 x 9.8

= 6860 N

c )

work done by friction for μ = .4

= .4 x 1400 x 9.8 x 107

= 587216 J

Initial Kinetic energy

= .5 x 1400 x 35 x 35

= 857500 J

Kinetic energy at the at of collision

= 857500 - 587216

= 270284 J

So , if v be the velocity at the time of collision

1/2 mv² = 270284

v = 19.65 m /s

d ) centripetal force required

= mv₀² / d which will be provided by frictional force

= (1400 x 35 x 35) / 107

= 16028 N

Maximum frictional force possible

= μmg

= .5 x 1400 x 9.8

= 6860 N

So this is not possible.

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3 years ago
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gregori [183]

Answer:

\lambda = 672 nm

so this is nearly red colour light

Explanation:

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\Delta x = 2\mu t + \frac{\lambda}{2}

now we know that for constructive interference of light the path difference is given as

\Delta x = n\lambda

so we will have

2\mu t + \frac{\lambda}{2} = N\lambda

so we will have

4\mu t = \lambda

\lambda = 2(1.40)(120nm)

\lambda = 672 nm

so this is nearly red colour light

8 0
3 years ago
An incident ray through the focal point of a spherical mirror will:
Delvig [45]
Reflect parallel of the principal axis
3 0
3 years ago
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