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kvv77 [185]
2 years ago
10

A microscope illuminator uses a transformer to step down the 120 V AC of the wall outlet to power a 12.0 V,50 W microscope bulb.

What is the resistance of the bulb filament
Engineering
1 answer:
Anna35 [415]2 years ago
5 0

Answer:2.88 ohms

Explanation:

R= V^2 / P

12^2/50

144/50

2.88 ohms

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Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 280
stiks02 [169]

Answer:

Minimum Allowable diameter for each of the 5 bolts = 0.0394 m = 3.94 cm

Explanation:

Maximum Working Stress = Ultimate Shear Stress/factor of safety

Maximum Working Stress = 280 MPa/3.8 = 73.68 MPa

Working Stress = Applied load/Minimum allowable Area = L/A

Minimum Allowable Area = Applied Load/Maximum Working Stress

A = 450000/73680000 = 0.00611 m²

This area is supplied by 5 bolts, so each bolt supplies A/5 = 0.0061/5 = 0.00122 m²

Cross sectional Area of bolts = πD²/4

0.00122 = πD²/4

D² = 4 × 0.00122/π = 0.00155

D = √0.00155 = 0.0394 m = 3.94 cm

Each of the five bolt can have a minimum diameter of 3.94 cm

Hope this Helps!!!

8 0
3 years ago
A 75 mm diameter, 15 mm wide A80-M12V grinding wheel is used for traverse surface grinding of a 75 mm by 75 mm 1040 steel workpi
Dmitrij [34]

Answer:

Detailed solution is given and calculations are done

3 0
3 years ago
A thin aluminum sheet is placed between two very large parallel plates that are maintained at uniform temperatures T1 = 900 K, T
Maru [420]

The net radiation heat transfer between the two plates per unit surface area of the plates with shield and without shied are respectively; 2282.76 W/m² and 9766.75 W/m²

<h3>How to find the net radiation heat transfer?</h3>

We are given;

Temperature 1; T₁

Temperature 2; T₂

Temperature 3; T₃

Emissivity 1; ε₁ = 0.3

Emissivity 2; ε₂ = 0.7

Emissivity 3; ε₃ = 0.2

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with shield is;

Q'₁₂ = σ(T₁⁴ - T₂⁴)]/[((1/ε₁) + (1/ε₂) - 1) + ((1/ε₃,₁) + (1/ε₃,₂) - 1)]

Q'₁₂ = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[((1/0.3) + (1/0.7) - 1) + ((1/0.15) + (1/0.15) - 1)]

Q'₁₂,shield = 2282.76 W/m²

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with no shield is;

Q'₁₂,no shield = σ(T₁⁴ - T₂⁴)]/((1/ε₁) + (1/ε₂) - 1))

Q'₁₂,no shield = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[(1/0.3) + (1/0.7) - 1)]

Q'₁₂,no shield = 9766.75 W/m²

Then the ratio of radiation heat transfer for the two cases becomes;

Q'₁₂,shield/Q'₁₂,no shield = 2282.76/9766.75 = 0.2337 or 4/17

Read more about Net Radiation Heat Transfer at; brainly.com/question/14148915

#SPJ1

8 0
2 years ago
The water level in a tank is 20 m above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end of
weeeeeb [17]

Answer:

P_{pump}= 68600Pa

Explanation:

We understand that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,

In this way,

P_{tank} + P_{pump} = P_{nozzle}

For the pressure we also know that it is given by

P=\rho gh

\rho gh_1 + P_{pump} = \rho gh_2

So,

P_{pump} = \rho g(h2 - h1)

P_{pump}= 1000*9.8(27-20)

P_{pump}= 68600Pa

7 0
3 years ago
State the two Carnot corollaries (principles)
Salsk061 [2.6K]

Answer  and Explanation:

The two principles or corollaries of Carnot Theorem are listed below:

1). The efficiencies of all the reversible heat engines between any two thermal reservoirs working between the same temperatures will be equal to each other.

2). For every Carnot engine working between any two thermal reservoirs will have the same efficiency independent of the operating conditions and the nature of working substance. It only depends on the temperature of the thermal energy reservoirs.

8 0
4 years ago
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