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navik [9.2K]
2 years ago
11

9. Calculate the total resistance and current in a parallel cir-

Engineering
1 answer:
Taya2010 [7]2 years ago
5 0

Answer:

  d. 2.3 ohms (5.3 amperes)

Explanation:

The calculator's 1/x key makes it convenient to calculate parallel resistance.

  Req = 1/(1/4 +1/8 +1/16) = 1/(7/16) = 16/7 ≈ 2.3 ohms

This corresponds to answer choice D.

__

<em>Additional comment</em>

This problem statement does not tell the applied voltage. The answer choices suggest that it is 12 V. If so, the current is 12/(16/7) = 21/4 = 5.25 amperes.

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What is the basic formula for actual mechanical advantage?
slavikrds [6]

Answer:

Mechanical Advantage Formula

The efficiency of a machine is equal to the ratio of its output to its input. It is also equal to the ratio of the actual and theoretical MAs. But, it does not mean that low-efficiency machines are of limited use. An automobile jack, for example, have to overcome a great deal of friction and therefore it has low efficiency. But still, it is extremely valuable because small effort can be applied to lift a great weight.

Also, in another way the mechanical advantage is the force generated by a machine to the force applied to it which is applied in assessing the performance of the machine.

The mechanical advantage formula is:

MA = FBFA

Explanation:

MAmechanical advantageFBthe force of the object

FAthe effort to overcome the force

3 0
2 years ago
Does somebody know how to do this?
maksim [4K]
No I don’t sorry, I hope you do well
4 0
2 years ago
Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb
guajiro [1.7K]

Answer:

\eta_{turbine} = 0.603 = 60.3\%

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = h_{g\ at\ 125KPa} = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than s_g and greater than s_f at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88

Now, we will find h_{2s}(enthalpy at the outlet for the isentropic process):

h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg

Now, the isentropic efficiency of the turbine can be given as follows:

\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%

3 0
3 years ago
Can you help me with this
Fiesta28 [93]

the function is to provide sealed combustion so that the loss of gas is minimized

6 0
1 year ago
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3 × 10-4
Vladimir [108]

Answer:

maximum stress is 2872.28 MPa

Explanation:

given data

radius of curvature = 3 × 10^{-4} mm

crack length = 5.5 × 10^{-2} mm

tensile stress = 150 MPa

to find out

maximum stress

solution

we know that  maximum stress formula that is express as

\sigma m = 2 ( \sigma o ) \sqrt{\frac{a}{\delta t}}     ......................1

here σo is applied stress and a is half of internal crack and t is radius of curvature of tip of internal crack

so put here all value in equation 1 we get

\sigma m = 2 ( \sigma o) \sqrt{\frac{a}{\delta t}}  

\sigma m = 2(150) \sqrt{ \frac{\frac{5.5*10^{-2}}{2}}{3*10^{-4}}}  

σm = 2872.28 MPa

so maximum stress is 2872.28 MPa

8 0
2 years ago
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