Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.

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Answer:

[tex]a=45.31m/s^2[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=5.74[/tex]

Force [tex]F=317N[/tex]

Gravitational Acceleration [tex]g=9.81m/s^2[/tex]

Generally the equation for Force is mathematically given by

 [tex]F-mg=ma[/tex]

 [tex]317-5.74*9.81=5.74 a[/tex]

 [tex]a=\frac{260.7}{5.74}[/tex]

 [tex]a=45.31m/s^2[/tex]

Answer:

a.  164 °F b. 91.11 °C c. 1439.54 kJ

Explanation:

a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?

Since the starting temperature is 48°F and the final temperature which water boils is 212°F, the number of degrees Fahrenheit we would need to raise the temperature is the difference between the final temperature and the initial temperature.

So, Δ°F = 212 °F - 48 °F = 164 °F

b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?

To find the degree change in Celsius, we convert the initial and final temperature to Celsius.

°C = 5(°F - 32)/9

So, 48 °F in Celsius is

°C₁ = 5(48 - 32)/9

°C₁ = 5(16)/9

°C₁ = 80/9

°C₁ = 8.89 °C

Also, 212 °F in Celsius is

°C₂ = 5(212 - 32)/9

°C₂ = 5(180)/9

°C₂ = 5(20)

°C₂ = 100 °C

So, the number of degrees in Celsius you must raise the temperature is the temperature difference between the final and initial temperatures in Celsius.

So, Δ°C = °C₂ - °C₁ = 100 °C - 8.89 °C = 91.11 °C

c. [2 pts] How much energy is required to heat the four quarts of water from

48°F to 212°F (boiling)?

Since we require 15.8 kJ for every degree Celsius of temperature increase of the four quarts of water, that is 15.8 kJ/°C and it rises by 91.11 °C, then the amount of energy Q required is Q = amount of heat per temperature rise × temperature rise =  15.8 kJ/°C × 91.11 °C = 1439.54 kJ

Answer:

a. unit of length: [L]

b. unit of volume: [[tex]L^3[/tex]]

c. unit of pressure:[tex]P=\frac{F}{A} \equiv\frac{[MLT^{-2}]}{[L^2]}[/tex] [tex][ML^{-1}T^{-2}][/tex]

d. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]

e. unit of force: [tex][kg.m/s^2]\equiv [MLT^{-2}][/tex]

f. unit of power: [tex]N.m.s^{-1}\equiv [ML^2T^{-3}][/tex]

Force: [tex]F=m.a=m.\frac{v}{t}=m.\frac{x}{t}\div t[/tex]

Power: [tex]P=\frac{W}{t}=\frac{F.x}{t}[/tex]

where:

F = force

A = area

W = work

t = time

a = acceleration

v = velocity

x = displacement

Solution :

Characteristic length  = thickness / 2

                                    [tex]$=\frac{0.04}{2}$[/tex]

                                    = 0.02 m

Thermal conductivity for steel is 42.5 W/m.K

[tex]$\text{Biot number} = \frac{\text{convective heat transfer coefficient} \times \text{characteristic length}}{\text{thermal conductivity}}$[/tex]

                  [tex]$=\frac{30 \times 0.02}{42.5}$[/tex]

                  = 0.014

Since the Biot number is less than 0.01, the lumped system analysis is applicable.

[tex]$\frac{T-T_{\infty}}{T_0-T_{\infty}} = e^{-b\times t}$[/tex]

Where,

T = temperature after t time

[tex]$T_{\infty}$[/tex] = surrounding temperature

[tex]$T_0$[/tex] = initial temperature

[tex]$b=\frac{\text{heat transfer coefficient}}{\text{density} \times {\text{specific heat } \times \text{characteristic length }}}$[/tex]

t = time

We calculate B:

[tex]$b=\frac{30}{7833 \times 460 \times 0.02}$[/tex]

  = 0.000416

Thus, [tex]$\frac{100-50}{500-50}=e^{-0.00416 \times t}$[/tex]

t = 5281.78 second

  = 88.02 minutes

Thus the time taken for reaching 100 degree Celsius is 88.02 minutes.

Answer: Can these nuts fit in your mouth?

Explanation:

im just here for the points >:)

Answer:

a) -505.229 kJ/Kg

b) -1.724 kJ/kg

Explanation:

T1 = 400°C

P1 = 3 MPa

P2 = 125 kPa

work output   = 530 kJ/kg

surrounding temperature = 20°C = 293 k

A) Calculate heat transfer from Turbine to surroundings

Q = h2 + w - h1

h ( enthalpy )

h1 = 3231.229 kj/kg

enthalpy at P2

h2 = hg = 2676 kj/kg

back to equation 1

Q = 2676 + 50 - 3231.229  = -505.229 kJ/Kg  ( i.e.  heat is lost )

b) Entropy generation

entropy generation = Δs ( surrounding )  + Δs(system)

                                =  - 505.229 / 293   + 0

                                = -1.724 kJ/kg  

Explanation:

This relationship, known as Faraday's law of induction (to distinguish it from his laws of electrolysis), states that the magnitude of the emf induced in a circuit is proportional to the rate of change of the magnetic flux that cuts across the circuit.

The values of Is and V are as: (a) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V. (b) [tex]Is = 4.56 \times 10^{-15} A[/tex] and V = 0.516 V. (c) [tex]Is = 1.18 \times 10^{-16} A\\[/tex] and V = 0.459 V. (d) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V.

The relation between the current and voltage of a diode is given by the Shockley diode equation. It is an exponential function and can be given by the following equation:

[tex]I = Is \times (e^{V/Vt} - 1)[/tex]

Where

(a)

Given that:

V = 0.700 V

And I = 1.00 A.

Substituting these values in the equation above to get,

[tex]1.00 A = Is \times (e^{0.700 V / 0.025 V} - 1)\\Is = 2.34 \times 10^{-11} A[/tex]

The terminal voltage at 10% of the measured current can be found by substituting I = 0.1 A in the above equation and solving for V as:

V = 0.581 V.

(b)

Given that:

V = 0.650 V

And, I = 1.00 mA.

Substituting these values in the equation above to get,

[tex]1.00 mA = Is \times (e^{0.650 V / 0.025 V} - 1)\\ Is = 4.56 \times 10^{-15} A[/tex]

The terminal voltage at 10% of the measured current can be found by substituting I = 0.1 mA in the above equation and solving for V as:

V = 0.516 V.

(c)

Given that:

V = 0.650 V

And, I = 10 μA.

Substituting these values in the equation above to get,

[tex]10 \mu A = Is \times (e^{0.650 V / 0.025 V} - 1)\\Is = 1.18 \times 10^{-16} A[/tex]

The terminal voltage at 10% of the measured current can be found by substituting I = 1 μA in the above equation and solving for V as:

V = 0.459 V.

(d)

Given that:

V = 0.700 V

And, I = 100 mA.

Substituting these values in the equation above to get,

[tex]100 \ mA = Is \times (e^{0.700 V / 0.025 V} - 1)\\Is = 2.34 \times 10^{-11} A[/tex]

The terminal voltage at 10% of the measured current can be found by substituting I = 10 mA in the above equation and solving for V as:

V = 0.581 V.

So, the values of Is and V are as: (a) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V. (b) [tex]Is = 4.56 \times 10^{-15} A[/tex] and V = 0.516 V. (c) [tex]Is = 1.18 \times 10^{-16} A\\[/tex] and V = 0.459 V. (d) [tex]Is = 2.34 \times 10^{-11} A[/tex] and V = 0.581 V.

Learn more about Terminal voltage here:

https://brainly.com/question/34372613

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Answer:

Input power = 465.63 W

current = 1.94 A

Explanation:

we have the following data to answer this question

V = 9130

i = 0.051

the input power = VI

I = 51.0 mA = 0.051

= 9130 * 0.051

= 465.63 watts

the current = 465.63/240

= 1.94A

therefore the input power is 465.63 wwatts

while the current is 1.94A

the input power is the same thing as the output power.

Solution :

Given :

[tex]$H(S) =\frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]

Transfer function, [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]

[tex]$Y(S) \left(S^2+\frac{10}{3}S+1\right) = (2S-2) \times (S)$[/tex]

[tex]$S^2Y(S) + \frac{10}{3}(SY(S)) + Y(S) = 2(S \times (S)) - 2 \times (S)$[/tex]

Apply Inverse Laplace Transforms,

[tex]$\frac{d^2y(t)}{dt^2} + \frac{10}{3} \frac{dy(t)}{dt} + y(t)=2 \frac{dx(t)}{dt} - 2x(t)$[/tex]

The above equation represents the differential equation of transfer function.

Given : [tex]$x(t)=e^{-t} u(t) \Rightarrow X(S) = \frac{1}{S+1}$[/tex]

We have : [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]

[tex]$Y(S) = X(S) \times \frac{6S-6}{3S^2+10 S + 3} = \frac{6S-6}{(S+1)(3S+1)(S+3)}$[/tex]

[tex]$Y(S) = \frac{A}{S+1}+\frac{B}{3S+1} + \frac{C}{S+3}[/tex]

[tex]$A = Lt_{S \to -1} (S+1)Y(S)=\frac{6S-6}{(3S+1)(S+3)} = \frac{-6-6}{(-3+1)(-1+3)} = 3$[/tex]

[tex]$B = Lt_{S \to -1/3} (3S+1)Y(S)=\frac{6S-6}{(S+1)(S+3)} = \frac{-6/3-6}{(1/3+1)(-1/3+3)} = \frac{-9}{2}$[/tex]

[tex]$C = Lt_{S \to -3} (S+3)Y(S)=\frac{6S-6}{(S+1)(3S+1)} = \frac{-18-6}{(-3+1)(-9+1)} = \frac{-3}{2}$[/tex]

So,

[tex]$Y(S) = \frac{3}{S+1} - \frac{9/2}{3S+1} - \frac{3/2}{S+3}$[/tex]

        [tex]$=\frac{3}{S+1} - \frac{3/2}{S+1/3} - \frac{3/2}{S+3}$[/tex]

Applying Inverse Laplace Transform,

[tex]$y(t) = 3e^{-t}u(t)-\frac{3}{2}e^{-t/3}u(t) - \frac{3}{2}e^{-3t} u(t)$[/tex]

       [tex]$=\frac{-3}{2}e^{-\frac{1}{3}t}u(t) + \frac{3}{1}e^{-t}u(t)-\frac{3}{2}e^{-3t} u(t)$[/tex]

where, a = 2

            b = 1

            c= 2

Answer:

I can't understand this language .

Answer:

1709.07 ft^3/s

Explanation:

Annual peak streamflow = Log10(Q [ft^3/s] )

mean = 1.835

standard deviation = 0.65

Probability of levee been overtopped in the next 15 years = 1/5

Determine the design flow ins ft^3/s

P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2

                         ∴  T = 67.72 years

Q₁₅ = 1 - 0.2 = 0.8

Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )

K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )

    = 2.1504

back to equation 1

Zt = 1.835 + ( 2.1504 * 0.65 )  = 3.23276

hence:

Log₁₀ ( Qt(ft^3/s) ) = Zt  = 3.23276

hence ; Qt = 10^3.23276

                  = 1709.07 ft^3/s

Answer:

The gage and absolute pressures are 900 and 1001 kilopascals, respectively.

Explanation:

The gage pressure ([tex]P_{g}[/tex]), in kilopascals, is the difference between absolute ([tex]P_{abs}[/tex]) and atmospheric pressures ([tex]P_{atm}[/tex]), measured in kilopascals. If we know that [tex]P_{g} = 900\,kPa[/tex] and [tex]P_{atm} = 101\,kPa[/tex], then the gage and absolute pressures are, respectively:

[tex]P_{g} = 900\,kPa[/tex]

[tex]P_{abs} = P_{atm} + P_{g}[/tex]

[tex]P_{abs} = 101\,kPa + 900\,kPa[/tex]

[tex]P_{abs} = 1001\,kPa[/tex]

The gage and absolute pressures are 900 and 1001 kilopascals, respectively.

Answer: hello your question is incomplete below is the complete question

answer:

N010 GO2 X7.0 Y2.0 15.0 J2.0  ( option 1 )

Explanation:

Given that the NC machining has to be moved from point ( 5,4 ) to point ( 7,2 ) along a circular path

GO2 = circular interpolation in a clockwise path

G91 = incremental dimension

hence the correct option is :

N010 GO2 X7.0 Y2.0 15.0 J2.0  

Answer:

[tex]h_{1} = h_2} = 293.45 KJ/kg[/tex].

The quality of the refrigerant exiting the expansion valve is

[tex]x_{2}=0.193596[/tex].

Explanation:

Fluid given Ammonia.

Inlet 1:-

Temperature [tex]T_{1}[/tex] = [tex]24^{o} C[/tex].

Pressure [tex]P_{1}[/tex] = 10 bar.

Exit 2:-

Pressure [tex]P_{2}[/tex] = 1 bar.

Solution:-

Answer:

The explanation according to the given query is summarized in the explanation segment below.

Explanation:

Answer:

Explanation:

#include<iostream>

using namespace std;

int main()

{

int n1,n2;

cout<<"Enter a number:"<<endl; //Entering first number

cin>>n1;

cout<<"Enter a number:"<<endl; //Entering second number

cin>>n2;

if(n1%2==0 && n1%2==0) //Checking whether the two number are even or not

{

if(n1>n2)

{

cout<<n1<<" is the larger number"<<endl;

}

else if(n1==n2)

{

cout<<"Numbers are equal"<<endl;

}

else

{

cout<<n2<<" is the larger number"<<endl;

}

}

else

{

cout<<"The number are not even"<<endl;

}

}