What is an actuator?

Voltage-regulated channels can be found a. at the motor end plate. b. on the surfaces of dendrites. c. in the membrane that cove

USPshnik [31]

Answer:

Option C

In the membrane that covers axons

Explanation:

Voltage-regulated channels allow for selective passage of different beneficial ions such as potassium and are found on the surface of a wide variety of cells such as nerve, muscle, and secretory cells. They mainly regulate cell membrane excitability, repetitive low frequency firing in some neurons, and recover the nerve fiber membrane.

7 0

3 years ago

Technician A says that weld-through primer can be removed from the immediate weld area to improve weld quality. Technician B say

lisabon 2012 [21]

Answer:

83737373777473737373738388383838

4 0

3 years ago

A 304 stainless steel (yield strength 30 ksi) cylinder has an inner diameter of 4 in and a wall thickness of 0.1 in. If it is su

Natasha_Volkova [10]

Answer:

The value we got is 1.423 ksi which is less than 30 and so the material hasn't failed and yielding hasn't occured.

Explanation:

This is a cylindrical thin walled vessel. Thus;

Hoop stress; σ1 = pr/t

Where ;

p is internal pressure

r_o is radius = 4/2 = 2 in

t is thickness of wall = 0.1 in

Thus;

σ1 = 70 x 2/0.1 = 1400 ksi

Longitudinal stress; σ2 = pr/2t

σ2 = 70 x 2/(0.1 x 2) = 700 ksi

Now, we want to find the normal stress. The inner radius of the circle will be; r_i = r_o - t = 2 - 0.1 = 1.9 in

So, normal stress by axial force is given by;

σ_fx = F/A = F/(π(r_o² - r_i²))

We are given that F = 500 lb

σ_fx = 500/(π(2² - 1.9²))

σ_fx = 408.1 ksi

We can now find the torsion from the formula;

τ = (T•r_o)/J

We are given that T = 70 lb.ft = 70 x 12 lb.in = 840 lb.in

J is the polar moment of inertia and has a formula; ((π/2)(r_o⁴ - r_i⁴))

So,J = ((π/2)(2⁴ - 1.9⁴)) = 4.662 in⁴

Thus,τ_xy = (840 x 2)/4.662 = 360.4 ksi

σ1 is in the y direction and σ2 is in the x direction. Thus;

σ_x = σ1 + σ_fx = 700 + 408.1 = 1108.1 ksi

Also, σ_y = σ1 = 1400 ksi

Now distortion energy can be expressed as;

σ_Y² = σ_x - σ_x•σ_y + (σ_y)² + 3(τ_xy)²

Plugging in the relevant values, we obtain ;

σ_Y² = 1108.1² - (1108.1*1400) + 1400² + 3(360.4)²

So, σ_Y² = 2.02 x 10^(6) psi²

σ_Y = √2.02 x 10^(6)

σ_Y = 1423 psi = 1.423 ksi

The question says the yield strength of the material is 30 ksi.

The value we got is less than 30 and so the material hasn't failed and yielding hasn't occured.

8 0

3 years ago

An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow

borishaifa [10]

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

$\\ E_{inflow} = E_{outflow}$

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy $\\ E_{inflow} = m_{1} h_{1}$ .

$\\$

$E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$

$\\$

Where $m_{1} h_{1}$ are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa $\\$ ,

$m_{2} h_{2}$ are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively. $\\$ ,

and $m_{3} h_{3}$ are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively. $\\$ ,

We can now express write out the required equation by substituting the new expression for the energies $\\$

$m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$

from the above equation, the unknown are the enthalpy values and the mass flow rate. $\\$

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table. $\\$

It is more convenient to start from outlet 3 were we have a temperature $100^{0}C$ and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have $\\$

$h_{3}$ = 2682.4 KJ/KG , at this point also from the table the entropy value , $s_{3}$ value is 7.6953 KJ/Kg.K. $\\$

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet. $\\$

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg. $\\$

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg. $\\$

Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e $m_{1} = m_{2} + m_{3}$ $\\$