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STatiana [176]
3 years ago
15

A thin 20-cm*20-cm flat plate is pulled at 1m/s horizontally through a 4-mm thick oil layer sandwiched between two stationary pl

ates,as shown in the figure.the dynamic viscosity of oil is 0.03pa.s .assuming the velocity in each layer to vary linearly, plot the velocity profile and explain why the profile takes that shape and determine the force that needs to be applied on the plate to mantain this motion when the h1=1.5mm and h2 2.5mm
Engineering
1 answer:
leva [86]3 years ago
3 0

Answer:

hindi ko polynomial alam

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A coil having resistance of 7 ohms and inductance of 31.8 mh is connected to 230v,50hz supply.calculate 1. The circuit current 2
lora16 [44]

(1) The current in the circuit is 18.87 A,

(2) The phase angle is 54.97°

(3) The power factor is 0.574

(4) The power consumed is 2491.2 W

(1) To calculate the current in the circuit, first, we need to find the overall impedance of the circuit.

We can calculate the overall impendence of the circuit using the formula below.

  • Z = √[R²+(2πfL)²]........................ Equation 1

Where:

  • R = resistance of the coil
  • f = Frequency
  • L = Inductance of the coil
  • Z = Overall impedance of the circuit

From the question,

Given:

  • R = 7 ohms
  • L = 31.8 mH = 0.0318 H
  • f = 50 Hz
  • π = 3.14

Substitute these values into equation 1

  • Z = √[7²+(2×3.14×50×0.0318)²]
  • Z = √(49+99.7)
  • Z = √(148.7)
  • Z = 12.19 ohms.

Therefore we use the formula below to calculate the current in the circuit.

  • I = V/Z.................. Equation 2

Where:

  • V = Voltage
  • I = current in the circuit.

Given:

  • V = 230 V.

Substitute into equation 2

  • I = 230/12.19
  • I = 18.87 A

(2) To calculate the phase angle, we use the formula below.

  • ∅ = tan⁻¹(2πfL/R)............... Equation 3

Where:

  • ∅ = Phase angle.


Substitute into equation 3

  • ∅ = tan⁻¹(2×3.14×50×0.0318/7)
  • ∅ = tan⁻¹(9.9852/7)
  • ∅ = tan⁻¹(1.426)
  • ∅ = 54.97°

(3) To calculate the power factor, we use the formula below.

  • pf = cos∅............ Equation 4

Where:

  • pf = power factor.

Substitute the value of ∅ into equation 4

  • pf = cos(54.97°)
  • pf = 0.574.

(4) And Finally to calculate the power consumed we use the formula below.

  • P = V×I×pf................ Equation 5

Where:

  • P = The power consumed

Substitute the values into equation 5

  • P = 230(18.87)(0.574)
  • P = 2491.22 W


Hence, (1) The current in the circuit is 18.87 A, (2) The phase angle is 54.97° (3) The power factor is 0.574 (4) The power consumed is 2491.2 W

Learn more about Impedance here: brainly.com/question/13134405

5 0
2 years ago
The ratio of the rate of etch-product formation to the flow rate of etch gas should be greater than 0.15 for uniform etching. Su
son4ous [18]

Answer:

answer for the question :

The ratio of the rate of etch-product formation to the flow rate of etch gas should be greater than 0.15 for uniform etching. Suppose a 200 mm diameter silicon wafer is etched at a rate of 50.0 nm/min in a CF4 plasma.(a) How many Si atoms are removed per minute?(b) What evolution rate of SiF4 does this correspond to in standard cubic centimeters per minute?(c) What minimum flow rate of CF4 should be maintained? "

is explained in the attachment.

Explanation:

7 0
4 years ago
What is the function and role of product tear down charts, and how do engineers utilize them in the reverse engineering process?
lora16 [44]

Answer:

Product Teardown 28 pieces (1) Plastic packaging: protect and display product for purchase. (4) Exterior screws: hold case halves together. (1) Right case half: acts as part of a handle and contains the rest of the parts. (1) Left case half: acts as part of a handle and contains the rest of the parts.

Explanation:

A product teardown process is an orderly way to know about a particular product and identify its parts, system functionality to recognize modeling improvement and identify cost reduction opportunities. Unlike the traditional costing method, tear down analysis collects information to determine product quality and price desired by the consumers.

7 0
3 years ago
A steady tensile load of 5.00kN is applied to a square bar, 12mm on a side and having a length of 1.65m. compute the stress in t
Shtirlitz [24]

Answer:

The stress in the bar is 34.72 MPa.

The design factor (DF) for each case is:

A) DF=0.17

B) DF=0.09

C) DF=0.125

D) DF=0.12

E) DF=0.039

F) DF=1.26

G) DF=5.5

Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

we will calculate the design stress in this case:

\displaystyle \sigma_{dis}=\frac{T_l}{Sup}=\frac{5.00KN}{(12\cdot10^{-3}m)^2}=34.72MPa

The design factor for metals is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}

The design factor for plastic and ceramics is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039

F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5

This material will break.

3 0
3 years ago
While supporting load P, the maximum contraction permitted in a 4.75-m long pipe column is 7 mm. Given that E =180 GPa for the c
Alexeev081 [22]

Answer:

\mathbf{stress \ \sigma = 264.6 \ Mpa}

Explanation:

From the concept of Hooke's  Law,

E =\dfrac{ stress \ \sigma}{ strain \  \varepsilon}

where;

strain \ \varepsilon = \dfrac{change \ in \ dimension }{original \ dimension}

strain \ \varepsilon = \dfrac{7 \ mm }{4.75 \times 10^{3} \ mm}

strain \ \varepsilon =0.00147368

Recall:

E =\dfrac{ stress \ \sigma}{ strain \  \varepsilon}

stress \ \sigma = E \times { strain \  \varepsilon}

stress \ \sigma = 180 \times 10^{3} \ Mpa \times 0.00147

\mathbf{stress \ \sigma = 264.6 \ Mpa}

Thus, the stress in the pipe at the maximum allowable contraction = 264.6 Mpa

7 0
3 years ago
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