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Leni [432]
3 years ago
9

Resistance in a circuit that has 35 mA of current and 5.6 V applied equals.

Engineering
1 answer:
natka813 [3]3 years ago
5 0

Answer:

R = 160 Ω

Explanation:

For this problem, we will simply apply Ohm's law to find the resistance in the circuit.  Ohm's law is as follows:

Voltage = Current x Resistance

We can solve the law for resistance:

Resistance = Voltage / Current

And now we just plug in our values:

Resistance = 5.6V / 35mA

Resistance = 5.6V / 0.035 A

Resistance = 160 Ω

Hence the resistance of the circuit is 160 Ω.

Cheers

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5. Bar stock of initial diameter = 90 mm is drawn with a draft = 10 mm. The draw die has an entrance angle = 18, and the coeffi
yulyashka [42]

Answer:

a)  0.2099

b)  46.5 MPa

c)  233765 N

d)  3896 W

Explanation:

a)

r = (A'' - A') / A'', where

A'' = 1/4 * π * D''²

A'' = 1/4 * 3.142 * 90²

A'' = 6362.55 mm²

D' = D'' - d = 90 - 10 = 80 mm

A' = 1/4 * π * D'²

A' = 1/4 * 3.142 * 80²

A' = 5027.2 mm²

r = (A'' - A') / A'

r = (6362.55 - 5027.2) / 6362.55

r = 1335 / 6362.55

r = 0.2099

b)

Draw stress = σd

Y' = k = 105 MPa

Φ = 0.88 + 0.12(D/Lc), where

D = 0.5 (90 + 80) = 85 mm

Lc = 0.5 [(90 - 80)/sin 18] = 16.18 mm

Φ = 0.88 + 0.12(85/16.18) = 1.51

σd = Y' * (1 + μ/tan α) * Φ * In(A''/A')

σd = 105 * (1 + 0.08/tan18) * 1.51 * In(6362.55/5027.2)

σd = 105 * 1.246 * 1.51 * 0.2355

σd = 46.5 MPa

c)

F = A' * σd

F = 5027.2 * 46.5

F = 233764.8 N

d)  

P = 233764.8 (1 m/min)

P = 233764.8 Nm/min

P = 3896.08 Nm/s

P = 3896.08 W

6 0
4 years ago
An offshore oil rig will drill into the ocean floor which is located 1 mile below the water surface. Determine the pressure at t
Luden [163]

Answer:

15.65 MPa 150 times of human body internal pressure

Explanation:

Given:

- Depth of the ocean for offshore drilling d = 1 mile

Find:

Determine the pressure at that location

What would happen to an unprotected, exposed person at that depth?

Solution:

- The pressure at a certain depth of a fluid can be calculated with:

                                       P = p_w*g*d

Where, P is the pressure , p_w is the density of water ( 997 kg / m^3 ).

- Hence @ d = 1.0 mile = 1.6 km = 1600 m:

                                       P = 997*9.81*1600

                                       P = 15.65 MPa

- Whereas the pressure inside a human body is 101 KPa, Pressure under ocean @  1 mile of depth is 150 times in magnitude, enough to crush the human body!

4 0
3 years ago
Two technicians are discussing cylinder- testing. Technician A says that when testing the power level of a specific cylinder usi
konstantin123 [22]

Answer:

The answer is "Both Technician A and Technician B".

Explanation:

The cylinder Testing is intended to assess locomotive inconsistency in CNS rodents, for example, whenever the animal moves within a transparent plastic tube, its preliminary activity is registered as it rises against the stadium wall.

In the given question both technicians are correct because both are reliable ways to check cylinders and the influence of the belief if every pathway has many more advantages than each other.

6 0
3 years ago
Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the
myrzilka [38]

Answer:

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}

The head associated with the pump is cleared:

h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})

Inlet and outlet velocities are found:

v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }

v_{1} \approx 4.278\,\frac{m}{s}

v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }

v_{2} \approx 11.573\,\frac{m}{s}

Now, the head associated with the pump is finally computed:

h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m

h_{pump} = 7.705\,m

The power that pump adds to the fluid is:

\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}

\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

4 0
3 years ago
You were recently hired by the Andrews CEO as a consultant to evaluate the performance of the Chief Financial Officer (CFO). As
goldfiish [28.3K]

Answer: Collect more information about the extent to which this reported behavior is impacting others

Explanation:

The options are:

a. Collect more information about the extent to which this reported behavior is impacting others.

b. Tell the CEO about what you have uncovered and ask for her permission to speak about it with the CFO.

c. Focus on the CFO’s work performance since his personal behavior is inappropriate in the workplace.

d. Inform the human resources department that there is a potential problem.

Given the information provided in the question, the most effective next step in the process will be to collect more information about the extent to which this reported behavior is impacting others. This will be vital in determining whether what is being said about the Chief Financial Officer is true or not and also to know if true, whether it has an affect on the performance of affected employees in the organization.

Option B is wrong as telling the CEO right away isn't the next step as adequate information must be gathered first.

Option C is also wrong as the focus shouldn't be on his work in the organization alone. His personal behavior should be addressed as well if he's wrong. A great organization is made up of great individuals with good behaviors.

Option D is also wrong as the next step should not be to inform the human resource.

Therefore, the correct option is A.

5 0
3 years ago
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