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3 years ago

14

If a master cylinder piston applies pressure to a brak piston that has twice as much square area what will be the force from the

second piston

1 answer:

sp2606 [1]3 years ago

5 0

Explanation:

Pressure = force/area

Pressure stays the same.

If the area is doubled the force is doubled.

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Calculate total hole mobility if the hole mobility due to lattice scattering is 50 cm2 /Vsec and the hole mobility due to ionize

Ad libitum [116K]

Answer:

The total hole mobility is 41.67 cm²/V s

Explanation:

Data given by the exercise:

hole mobility due to lattice scattering = ul = 50 cm²/V s

hole mobility due to ionized impurity = ui = 250 cm²/V s

The total mobility is equal:

$\frac{1}{u} =\frac{1}{ul} +\frac{1}{ui} \\\frac{1}{u}=\frac{1}{50} +\frac{1}{250} \\u=41.67cm^{2} /Vs$

5 0

3 years ago

Read 2 more answers

The spring has a stiffness k = 200 N>m and an unstretched length of 0.5 m. If it is attached to the 3-kg smooth collar and th

ruslelena [56]

Answer:

15.467 m/s

Explanation:

See attached picture.

4 0

4 years ago

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be

cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, $\zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )$

Creep rate at 800⁰C, $\zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\$

$0.01 = C \exp(\frac{-Q}{1073R} )$ .........................(1)

Creep rate at 700⁰C

$\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2} \% per hour =5.5 * 10^{-4}$

$5.5 * 10^{-4} = C \exp(\frac{-Q}{1073R} )$ .................(2)

Divide equation (1) by equation (2)

$\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\$

Take the natural log of both sides

$ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol$

3 0

3 years ago

During a load test, a battery's voltage drops below a specific value. what action should the technician take?

Katyanochek1 [597]

Answer:

B

Explanation:

allow battery to become fully discharged and retest

4 0

3 years ago

Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire

Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

7 0

3 years ago