All categories

2 years ago

Which of these cars traveled faster during time interval please show solution

Engineering

1 answer:

Mazyrski [523]2 years ago

3 0

Answer:

I think D is correct

Explanation:

C is decreasing function, probably worst

A is arctan -> in radian, the rate of increasing is very slow-> second worst

B(14) = ln(9*14) = 4.8

D(14) = sqrt(8+14^2)=14.2

You might be interested in

Cynthia is producing a sculpture using material introduced in the Bronze Age. What two metals is she mixing?

Airida [17]

The correct answer
would be d
Iron and carbon
hope this helps

5 0

2 years ago

A 0.19-m3 rigid tank equipped with a pressure regulator contains steam at 2 MPa and 300°C. The steam in the tank is now heated.

AVprozaik [17]

Answer:

576.21kJ

Explanation:

#We know that:

The balance mass $m_{in}+m_{out}=\bigtriangleup m_{system}$

so, $m_e=m_1-m_2$

$Energy \ Balance\\E_{in}-E_{out}=\bigtriangleup E_{system}\\\\\therefore Q_i_n+m_eh_e=m_2u_2-m_1u_1$

#Also, given the properties of water as;

$(P_1=2Mpa,T_1=300\textdegree C)->v_1=0.12551m^3/kg,u_1=2773.2kJ/kg->h_1=3024.2kJ/kg\\\\(P_2=2Mpa,T_1=500\textdegree C)->v_2=0.17568m^3/kg,u_1=3116.9kJ/kg->h_1=3468.3kJ/kg$

#We assume constant properties for the steam at average temperatures: $h_e=\approx(h_1+h_2)/2$

#Replace known values in the equation above; $h_e=(3024.2+3468.3)/2=3246.25kJ/kg\\\\m_1=V_1/v_1=0.19m^3/(0.12551m^3/kg)=1.5138kg\\\\m_2=V_2/v_2=0.19m^3/(0.17568m^3/kg)=1.0815kg$

#Using the mass and energy balance relations;

$m_e=m_1-m_2\\\\m_e=1.5138-1.0815\\\\m_e=0.4323kg$

#We have $Q_i_n+m_eh_e=m_2u_2-m_1u_1$ : we replace the known values in the equation as;

$Q_i_n+m_eh_e=m_2u_2-m_1u_1\\\\Q_i_n=0.4323kg\times3246.2kJ/kg+1.0815kg\times3116.9-1.5138kg\times2773.2kJ/kg\\\\Q_i_n=573.21kJ$

#Hence,the amount of heat transferred when the steam temperature reaches 500°C is 576.21kJ

5 0

2 years ago

What is Back EMF? How does it limits the speed of a permanent magnet DC?

ss7ja [257]

Answer and Explanation:

The DC motor has coils inside it which produces magnetic field inside the coil and due to thus magnetic field an emf is induced ,this induced emf is known as back emf. The back emf always acts against the applied voltage. It is represented by $E_b$

The back emf of the DC motor is given by $E_b=\frac{NP\Phi }{60A}$

Here N is speed of the motor ,P signifies the number of poles ,Z signifies the the total number of conductor and A is number of parallel paths

As from the relation we can see that back emf and speed ar dependent on each other it means back emf limits the speed of DC motor

8 0

2 years ago

A heat pump operates on a vapor-compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the com

Rudiy27

Answer:

Hello your question has some missing information below are the missing information

The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump

answer : 2.49

Explanation:

For vapor-compression refrigeration cycle

P1 = P4 ; P1 = 140 kPa

P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa

<u>From pressure table of R 134a refrigerant</u>

h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg

h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )

= 296.8kJ/kg

h3 ( enthalpy of saturated liquid at P3 = 800 kPa ) = 95.47 kJ/kg

also h4 = 95.47 kJ/kg

To determine the coefficient of performance

Cop = ( h1 - h4 ) / ( h2 - h1 )

∴ Cop = 2.49

3 0

2 years ago

Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 450 K, 350 kPa, and velocity of 3

viktelen [127]

Answer:

Given that

Mass flow rate ,m=2.3 kg/s

T₁=450 K

P₁=350 KPa

C₁=3 m/s

T₂=300 K

C₂=460 m/s

Cp=1.011 KJ/kg.k

For ideal gas

P V = m R T

P = ρ RT

$\rho_1=\dfrac{P_1}{RT_1}$

$\rho_1=\dfrac{350}{0.287\times 450}$

ρ₁=2.71 kg/m³

mass flow rate

m= ρ₁A₁C₁

2.3 = 2.71 x A₁ x 3

A₁=0.28 m²

Now from first law for open system

$h_1+\dfrac{C_1^2}{200}+Q=h_2+\dfrac{C_2^2}{2000}$

For ideal gas

Δh = CpΔT

by putting the values

$1.011\times 450+\dfrac{3^2}{200}+Q=1.011\times 300+\dfrac{460^2}{2000}$

$Q=1.011\times 300+\dfrac{460^2}{2000}-\dfrac{3^2}{200}-1.011\times 450$

Q= - 45.49 KJ/kg

Q =- m x 45.49 KW

Q= - 104.67 KW

Negative sign indicates that heat transfer from air to surrounding

4 0

2 years ago