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3 years ago

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The selective thickness shims are measured with a(n): Group of answer choices hydraulic press. outside micrometer. dial indicato

r. depth micrometer.

1 answer:

alexgriva [62]3 years ago

7 0

The selective thickness shims are measured with an outside micrometer.

<h3>What is a micrometer?</h3>

It should be noted that a micrometer simply means an instrument for making precise linear measurement of dimensions.

In this case, the selective thickness shims are measured with an outside micrometer.

Learn more about micrometer on:

brainly.com/question/2176082

#SPJ12

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Rope BCA passes through a pulley at point C and supports a crate at point A. Rope segment CD supports the pulley and is attached

djverab [1.8K]

Answer:

363 pounds 32 degrees

Explanation:

Express your answers numerically in pounds and degrees to three significant figures separated by a comma. slader

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3 years ago

Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held

77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} }$

${T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} } = 280.15 \times \left (9 \right )^{\frac{1.333-1}{1.333} } = 485.03\ K$

The heat supplied = $\dot {m}_f$ × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = $\dot m$ · $c_p(T_3 - T_2)$

$\dot m$ = 20 kg/s

The heat supplied = 20* $c_p(T_3 - T_2)$ = 21,350 kJ/s

$c_p$ = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

$\dfrac{T_5}{T_4} = \dfrac{2}{1.333 + 1}$

T₅ = 1208.42*(2/2.333) = 1035.94 K

$C_j = \sqrt{\gamma \times R \times T_5}$ = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = $\dot m$ × $C_j$ = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

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3 years ago

I NEED ANSWER'S ASAP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

nydimaria [60]

Answer:

Sealing agents are made of eco-friendly materials that are difficult to properly position in hard-to-reach places.

Explanation:

4 0

4 years ago

Read 2 more answers

an electric circuit includes a voltage source and two resistances (50 and 75) in parallel. determine the voltage source required

ASHA 777 [7]

Answer:

The voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

Explanation:

Given;

Resistance, R₁ = 50Ω

Resistance, R₂ = 75Ω

Total resistance, R = (R₁R₂)/(R₁ + R₂)

Total resistance, R = (50 x 75)/(125)

Total resistance, R = 30 Ω

According to ohms law, sum of current in a parallel circuit is given as

I = I₁ + I₂

$I = \frac{V}{R_1} + \frac{V}{R_2}$

Voltage across each resistor is the same

$1.6 = \frac{V}{R_2}$

V = 1.6 x R₂

V = 1.6 x 75

V = 120 V

Therefore, the voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

This voltage is also the same for 50 ohms resistance but the current will be 2.4 A.

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3 years ago

Read 2 more answers

An automotive fuel cell consumes fuel at a rate of 28m3/h and delivers 80kW of power to the wheels. If the hydrogen fuel has a h

EastWind [94]

Answer:

The efficiency of this fuel cell is 80.69 percent.

Explanation:

From Physics we define the efficiency of the automotive fuel cell ( $\eta$ ), dimensionless, as:

$\eta = \frac{\dot W_{out}}{\dot W_{in}}$ (Eq. 1)

Where:

$\dot W_{in}$ - Maximum power possible from hydrogen flow, measured in kilowatts.

$\dot W_{out}$ - Output power of the automotive fuel cell, measured in kilowatts.

The maximum power possible from hydrogen flow is:

$\dot W_{in} = \dot V\cdot \rho \cdot L_{c}$ (Eq. 2)

Where:

$\dot V$ - Volume flow rate, measured in cubic meters per second.

$\rho$ - Density of hydrogen, measured in kilograms per cubic meter.

$L_{c}$ - Heating value of hydrogen, measured in kilojoules per kilogram.

If we know that $\dot V = \frac{28}{3600}\,\frac{m^{3}}{s}$ , $\rho = 0.0899\,\frac{kg}{m^{3}}$ , $L_{c} = 141790\,\frac{kJ}{kg}$ and $\dot W_{out} = 80\,kW$ , then the efficiency of this fuel cell is:

(Eq. 1)

$\dot W_{in} = \left(\frac{28}{3600}\,\frac{m^{3}}{s}\right)\cdot \left(0.0899\,\frac{kg}{m^{3}} \right)\cdot \left(141790\,\frac{kJ}{kg} \right)$

$\dot W_{in} = 99.143\,kW$

(Eq. 2)

$\eta = \frac{80\,kW}{99.143\,kW}$

$\eta = 0.807$

The efficiency of this fuel cell is 80.69 percent.

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3 years ago