banana with an average mass of 0.15 kg and average specific heat of 3.35 kJ/kg · °C is cooled from 20°C to 5°C. The amount of heat transferred from the banana is

a.
62.1 kJ

b.
7.5 kJ

c.
None of these

d.
6.5 kJ

e.
0.85 kJ

f.
17.7 kJ

Answer:

a) at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

b) daylight (d) = 0.50 μm

    Incandescent ( i ) =  1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

Values are gotten from the table named: blackbody radiation functions

a) Calculate the band emission fractions for the visible region

at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

attached below is a detailed solution to the problem

b)calculate wavelength corresponding to the maximum spectral intensity

For daylight ( d ) = 2898 μm *k / 5800 k  = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

Answer:

D

Explanation:

Got it wrong so i could answer

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

([tex]p_t[/tex])₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [tex][[/tex] α₂/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex] ------- let this be equation 2

where ([tex]p_t[/tex])₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex]  = 2(6( σ₀ )₁) [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]  

divide both sides by 2(σ₀)₁

[tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex]  =  6 [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]

[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex]  =  6 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]

[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]][/tex]  =  36 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]][/tex]

1 / ([tex]p_t[/tex])₁ = 30.24 / ([tex]p_t[/tex])₂

([tex]p_t[/tex])₂ = 30.24([tex]p_t[/tex])₁

([tex]p_t[/tex])₂/([tex]p_t[/tex])₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

Answer:

1.4 psi

Explanation:

Before diving into the solution to the question above, let's pick out the parameters needed in solving this problem from the question.

=> The measurement for the outer diameter of the balloon = 20 inches, the measurement for the thickness =  0.012 in,  the allowable tensile stress = 1ksi and the allowable shear stress in the balloon =  0.3 ksi.

The first thing to determine is the inner diameter = 20 - 2 ×  0.012 in = 19.976 in.

Therefore, the tensile stress:

1000 = k × [19.976/2]÷ 2 × 0.012 = 2.4 psi.

Also, the sheer stress which is also the maximum permissible pressure in the balloon can be calculated below as:

0.3 × 1000 = k × [19.976/2]/ 4 × 0.012 = 1.4 psi.

Explanation:

D the answer is D and anyone can answer this Question

Based on the provided options, the scientific expression that is the same as 1545.5347 is 154553.4e3. Therefore, the correct option is option D.

A scientific expression, often known as scientific notation, is a method of representing extremely big or extremely small integers. It is often used to simplify the representation of such numbers in scientific and mathematical operations. A number is expressed in scientific notation as a product of two parts: a coefficient and a power of ten.

The coefficient is a value between 1 and 10, and the power of 10 determines how far the decimal point should be moved. The "e3" signifies multiplying the integer by 10 raised to the power of 3, which is similar to moving the decimal point three places to the right in this formula. As a result, 154553.4e3 equals 154553.4103, which simplifies to 154553400.

Therefore, the correct option is option D.

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Answer:

a) converting Martensite to spheroidite

The heat treatment procedure for converting Martensite to spheroidite  involves heating Martensite of  0.76 wt% C steel for approximately 24 hours at a temperature of 700°C

b) Converting Spheroidite to martensite

The heat treatment entails heating Spheroidite of 0.76 wt% C steel to a temperature of 760°C  to austenization then it will  be quenched at temperature > 140°C

c) Converting Bainite to Pearlite

The heat treatment involves  heating Bainite of 0.76 wt% C steel to a temperature of 720°C  until austenization then it will  be quenched at temperature < 35°C

d) Converting Pearlite to Bainite

The heat treatment involves heating Pearlite of  0.76 wt% C steel to a temperature of 720°C until austenization then it will  be cooled at a temperature range 220°C to 540°C. the temperature is maintained at this range until the complete formation of Bainite

e) Converting Spheroidite to perlite

The heat treatment involves  heating Spheroidite of 0.76 wt% C steel to a temperature of 720°C  until austenization then it will  be quenched at temperature < 35°C

f) Perlite to Spheroidite

The heat treatment procedure for converting Perlite to spheroidite  involves heating Perlite of  0.76 wt% C steel for approximately 24 hours at a temperature of 700°C

g) Tempered martensite to martensite

The heat treatment entails heating Tempered martensite   of 0.76 wt% C steel to a temperature of 760°C  until austenization then it will  be quenched at temperature > 140°C

h) Bainite to spheroidite

The heat treatment procedure for converting Perlite to spheroidite  involves heating Bainite of  0.76 wt% C steel for approximately 24 hours at a temperature of 700°C

Explanation:

The heat treatment procedure is simply the heating of a metal to a high temperature  and cooling the metal back. during this process the metal will undergo certain mechanical changes

a) converting Martensite to spheroidite

The heat treatment procedure for converting Martensite to spheroidite  involves heating Martensite of  0.76 wt% C steel for approximately 24 hours at a temperature of 700°C

b) Converting Spheroidite to martensite

The heat treatment entails heating Spheroidite of 0.76 wt% C steel to a temperature of 760°C  to austenization then it will  be quenched at temperature > 140°C

c) Converting Bainite to Pearlite

The heat treatment involves  heating Bainite of 0.76 wt% C steel to a temperature of 720°C  until austenization then it will  be quenched at temperature < 35°C

d) Converting Pearlite to Bainite

The heat treatment involves heating Pearlite of  0.76 wt% C steel to a temperature of 720°C until austenization then it will  be cooled at a temperature range 220°C to 540°C. the temperature is maintained at this range until the complete formation of Bainite

e) Converting Spheroidite to perlite

The heat treatment involves  heating Spheroidite of 0.76 wt% C steel to a temperature of 720°C  until austenization then it will  be quenched at temperature < 35°C

f) Perlite to Spheroidite

The heat treatment procedure for converting Perlite to spheroidite  involves heating Perlite of  0.76 wt% C steel for approximately 24 hours at a temperature of 700°C

g) Tempered martensite to martensite

The heat treatment entails heating Tempered martensite   of 0.76 wt% C steel to a temperature of 760°C  until austenization then it will  be quenched at temperature > 140°C

h) Bainite to spheroidite

The heat treatment procedure for converting Perlite to spheroidite  involves heating Bainite of  0.76 wt% C steel for approximately 24 hours at a temperature of 700°C

Answer:

letse see

Explanation:

Answer:

Va / Vb = 0.5934

Explanation:

First step is to determine total head losses at each pipe

at Pipe A

For 1/4 open gate valve head loss = 17 *Va^2 / 2g

elbow loss = 0.75 Va^2 / 2g

at Pipe B

For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g

elbow loss = 0.75 * Vb^2 / 2g

Given that both pipes are parallel

17 *Va^2/2g +  0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g  + 0.75 * Vb^2 / 2g

∴ Va / Vb = 0.5934



Explanation:

Answer:

L = spindle

M = lower ball joint

part without the letter showing = steering knuckle

Explanation:

Answer:

True

Explanation:

During expansion process, the boundary work is positive while in case of contraction, the boundary work is negative. During expansion process, the work is done by the system while in case of compression process, work in done on the system

Hence, the given statement is true

Answer:

3  kmol/m^3

Explanation:

Determine the concentration of C in the stream leaving the reactor

Given that the CTSR reaction ; A (k1) → B (k2) → C

K1 = 2 min^-1 , K2 = 3 min^-1 , time constant ; τ = V/F.= 0.5 min also n1 = n2

attached below is the detailed solution

concentration of C leaving the reactor= 3 kmol/mol^3

Given ; Ca = 5 kmol/m^3 , Cb = 2 kmol/m^3 ( from the attached calculations ) Cc = 3 kmol/m^3

Complete question:

If it took a 30m³ capacity tank mediated by 2cm waterproof water faucet for 10 hours, calculate the flow speed (exit) of water from the tank.

Answer:

the flow rate of the water from the tank is 0.05 m³/min

Explanation:

Given;

volume of water in the tank, v = 30 m³

length of the waterproof faucet, L = 2cm = 0.02 m

duration of water flow through the tank, t = 10 hours

The flow rate of the water from the tank is calculated as;

[tex]Q = \frac{V}{t} = \frac{30 \ m^3}{10\ h \ \times \ 60 \min} = 0.05 \ m^3/ \min[/tex]

Therefore, the flow rate of the water from the tank is 0.05 m³/min

Answer: hello your question lacks some data below is the missing data

Air at 32C has H = 0.204 kJ/mol and at 50C has H = -0.576 kJ/mol

H of steam can be found on the steam tables – vapor at 32C and 1 atm; vapor at 5C and liquid at 5C. Assume the volume of the humid air follows the ideal gas law.

H of water liquid at 5C = 21 kJ/kg; vapor at 5C = 2510.8 kJ/kg; H of water vapor at 32C = 2560.0 kJ/kg

Answer :

a) 34.98 lit/min

b) 1432.53 m^3/min

Explanation:

a) Calculate how much water is produced

density of water = 1 kg/liter

First we will determine the mass of condensed water using the relation below

inlet mass - outlet vapor mass =  0.0339508 * n * 18/1000 ----- ( 1 )

where : n = 57241.57

hence equation 1 = 34.98 Kg/min

∴ volume of water produced =  mass of condensed water / density of water

                                                =  34.98 Kg/min / 1 kg/liter

                                                = 34.98 lit/min

b) calculate the Volumetric flow rate of air entering the unit

applying the relation below

Pv = nRT

101325 *V = 57241.57 * 8.314 * 305  

∴ V = 1432.53 m^3/min

Answer: Power

Explanation:

The rate at which work is done or the amount of work done based on a period of time is referred to as power.

Power can also be defined as the amount of energy that is being transferred per unit time. The unit of power is one joule per second or simply called the watt.

Answer:

the surface temperature of the plates when they come out of the oven is approximately 445 °C

Explanation:

Given the data in the question;

thickness t = 3 cm = 0.03 m

so half of the thickness L = 0.015 m

thermal conductivity of brass k = 110 W/m°C

Density p = 8530 kg/m³

specific heat [tex]C_p[/tex] = 380 J/kg°C

thermal diffusivity of brass ∝ = 33.9 × 10⁻⁶ m²/s

Temperature of oven T₀₀ = 700°C

initial temperature T[tex]_i[/tex] = 25°C

time t = 10 min = 600 s

convection heat transfer coefficient h = 80 W/m².K

Since the plate is large compared to its thickness, the heat conduction is one dimensional. heat transfer coefficient and thermal properties are constant over the entire surface.

So, using analytical one-term approximation method, the Fourier number > 0.2.

now, we determine the Biot number for the process

we know that; Biot number Bi =  hL / k

so we substitute

Bi =  hL / k

Bi = (80 × 0.015) / 110 = 1.2 / 110 = 0.0109

Now, we get the constants λ₁ and A₁ corresponding to Biot Number ( 0.0109 )

The interpolation method used to find the

λ₁ = 0.1039 and A₁ = 1.0018

so

The Fourier number т = ∝t/L²

we substitute

Fourier number т = ( (33.9 × 10⁻⁶)(600) ) / (0.015)²

т = 0.02034 / 0.000225

т = 90.4

As we can see; 90.4 > 0.2

So,  analytical one-term approximation can be used.

∴ Temperature at the surface will be;

θ(L,t)[tex]_{wall[/tex] = (T(x,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀) ----- let this be equation

θ(L,t)[tex]_{wall[/tex] = [A₁e^(-λ₁²т)]cos( λ₁L / L )

so we substitute

θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- (0.1039)²× 90.4 )] cos( 0.1039 × 0.015 / 0.015 )

θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- 0.975886984 )] cos( 0.1039 )

θ(L,t)[tex]_{wall[/tex] = [1.0018 × 0.376857938] × 0.999998

θ(L,t)[tex]_{wall[/tex] = 0.3775

so we substitute into equation 1

θ(L,t)[tex]_{wall[/tex] = (T(L,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀)

0.3775 = ( T(L,t) - 700 ) / ( 25 - 700 )

0.3775 = ( T(L,t) - 700 ) / ( - 675 )

0.3775 × ( - 675 ) = ( T(L,t) - 700 )

- 254.8125 = T(L,t) - 700

T(L,t) = 700 - 254.8125

T(L,t) = 445.1875 °C ≈ 445 °C

Therefore, the surface temperature of the plates when they come out of the oven is approximately 445 °C

Answer:

D

Explanation:

All of the above

Answer:

Engine

Explanation:

Semi-independent suspension is the most common type of suspension system used on body over frame vehicles.

Semi-independent suspension give the front wheels some individual movement.

This suspension only used in rear wheels.

Thus, the correct option is Semi-independent suspension

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Answer:

the liquid woulriekwvhrnsshsnekwb ndrhwmoadi