Select the true statements regarding rigid bars. a. A rigid bar can bend but does not change length.
b. A rigid bar does not bend regardless of the loads acting upon it.
c. A rigid bar deforms when experiencing applied loads.
d. A rigid bar is unable to translate or rotate about a support.
e. A rigid bar represents an object that does not experience deformation of any kind.

Answers

Answer 1

Answer:

Answer:

option b and E are true

Explanation:

A lever is an example of a rigid bar that can rotate around a given point. In a rigid material, the existing distance does not change whenever any load is placed on it. In such a material, there can be no deformation whatsoever. Wit this explanation in mind:

option a is incorrect, given that we already learnt that no deformation of any kind happens in a rigid bar.

option b is true. A rigid bar remains unchanged regardless of the load that it carries.

option c is incorrect, a rigid bar does not deform with loads on it

option d is incorrect. A lever is a type of rigid bar, a rigid bar can rotate around a support.

option e is true. A rigid bar would not experience any deformation whatsoever.

Answer:

24Ω

Explanation:

When resistors are connected in parallel, the reciprocal of their combined resistance, when read with a DMM (Digital Multimeter - for measuring various properties of a circuit or circuit element such as resistance...) is the sum of the reciprocals of their individual resistances.

For example if two resistors of resistances R₁ and R₂ are connected together in parallel, the reciprocal of their combined resistance Rₓ is given by;

$(1)/(R_x)$ = $(1)/(R_1)$ + $(1)/(R_2)$

Solving for Rₓ gives;

$R_(x)$ = $(R_1 * R_2)/(R_1 + R_2)$ ------------------(i)

From the question;

Let

R₁ = resistance of first resistor = 40Ω

R₂ = resistance of second resistor = 60Ω

Now,

To get their combined or total resistance, Rₓ, substitute these values into equation (i) as follows;

$R_(x)$ = $(40 * 60)/(40 + 60)$

$R_(x)$ = $(2400)/(100)$

$R_(x)$ = 24 Ω

Therefore, the total resistance is 24Ω

Answer:

24 ohms

Explanation:

hope this helped :)

God bless you!!!!

1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty percent of the flow is extracted at 1000 kPa to a feedwater heater and the remainder flows out at 200 kPa. Find the two exit temperatures and the turbine power output.

Answers

Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

h2=3116KJ/kg

T2=Temperature(Water;P=P2;s=s1)

T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

W=m1(h1-h2)+m3(h2-h3)

W=5(3457-3116)+4(3116-2750)=3168Kw

An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a lagging power factor of 0.77. Determine the size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging.

Answers

Answer:

$Q=41.33 KVAR\ \nat\n\ 480 Vrms$

Explanation:

From the question we are told that:

Voltage $V=480/0 \textdegree V$

Power $P=120kW$

Initial Power factor $p.f_1=0.77 lagging$

Final Power factor $p.f_2=0.9 lagging$

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

$p.f_1=0.77$

$cos \theta_1 =0.77$

$\theta_1=cos^(-1)0.77$

$\theta_1=39.65 \textdegree$

And

$p.f_2=0.9$

$cos \theta_2 =0.9$

$\theta_2=cos^(-1)0.9$

$\theta_2=25.84 \textdegree$

Therefore

$Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=-41.33VAR$

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

$Q=41.33 KVAR\ \nat\n\ 480 Vrms$

What is 29.95 inHg in kPa?

Answers

Answer:

101.42235 kPa

Explanation:

The unit inHg means "inches of mercury", Its a pressure unit commonly used by the US aviators.

The conversion value to KPa (kilopascal) is

1 inHg= 3.386389 kPa

So now we only have to multiply:

29.95 inHg * 3.386389 kPa/in Hg =101.42235 kPa

Have a nice day and Good Luck!

What are the important factors needed to be considered while selecting a brake or clutch?

Answers

Answer:

The correct answer is: the following factors are needed to properly consider while selecting a brake or clutch:

-Engagement

-Friction

-Electromagnetic

-Mechanical

-Actuation

-Electric

-Fluid power

-Self-actuation

-Key concepts

-Application notes

-Selection criteria

Explanation:

Clutches and brakes are important devices in many rotating drive systems, it is very important to guarantee the security and the proper function of them accomplishing a high quality parameters in those factors.

1.19. A gas is confined in a 0.47 m diameter cylinder by a piston, on which rests a weight. The mass of the piston and weight together is 150 kg. The local acceleration of gravity is 9.813 m·s−2, and atmospheric pressure is 101.57 kPa. (a) What is the force in newtons exerted on the gas by the atmosphere, the piston, and the weight, assuming no friction between the piston and cylinder? (b) What is the pressure of the gas in kPa? (c) If the gas in the cylinder is heated, it expands, pushing the piston and weight upward. If the piston and weight are raised 0.83 m, what is the work done by the gas in kJ? What is the change in potential energy of the piston and weight?\

Answers

In this exercise we have to use the knowledge of force to calculate the required energies, so we have to:

a) 19094 N

b) 110.055 kPa

c) 1222 J

What is the concept of force?

In the field of physics, force is a physical action that causes deformation or that changes the state of rest or movement of a given object.

a) Knowing that the force formula is defined by:

$F = P + p * A\nF = m * g + p *\pi /4 * d^2\nF = 150 * 9.813 + 101570 * \pi /4 * 0.47^2 = 19094 N$