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3 years ago

What should be given to a customer before doing a repair?

Engineering

1 answer:

natima [27]3 years ago

7 0

A. I believe, lmk if I’m right

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Led test lights are used to test circuits that include controllers and computers. True or false

Marianna [84]

Answer:

True

Explanation:

An LED test light is a piece of electronic test equipment used to determine the presence of electricity in a piece of equipment under test, making this statement true.

3 0

2 years ago

A design team is working on creating a new locker organizer. They have

astraxan [27]

A. Present a Solution

8 0

4 years ago

In Process Costing, the journal entry to record the cost of goods transferred out from the WIP shaping department to the WIP pac

ddd [48]

Answer:

Journal entry :

Date Account and explanation Debit credit

WIP-Packing department XXX

WIP-Shaping department XXX

(To record cost of goods transferred out from the WIP shaping department to the WIP packaging department )

So above statement is False

Explanation:

4 0

3 years ago

The driveshaft of an automobile is being designed to transmit 134 hp at 2900 rpm. Determine the minimum diameter d required for

Misha Larkins [42]

Answer:

The minimum diameter is 1.344 in

Explanation:

The angular speed of the driveshaft is equal to:

$w=\frac{2\pi N}{60}$

Where

N = rotational speed of the driveshaft = 2900 rpm

$w=\frac{2\pi *2900}{60} =303.69rad/s$

The torque in the driveshaft is equal to:

$\tau=\frac{P}{w}$

Where

P = power transmitted by the driveshaft = 134 hp = 73700 lb*ft/s

$\tau=\frac{73700}{303.69} =242.68lb*ft$

The minimum diameter is equal to:

$d_{min} =(\frac{16T}{\pi *\tau } )^{1/3}$

Where

T = shear stress = 6100 psi

τ = 242.68 lb*ft = 2912.16 lb*in

$d_{min} =(\frac{16*2912.16}{\pi *6100} )^{1/3} =1.344in$

4 0

3 years ago

A fluid with a relative density of 0.9 flows in a pipe which is 12 m long and lies at an angle of 60° to the horizontal At the t

Minchanka [31]

Answer:

$Q=7.3\times 10^{-3} m^3/s$

Explanation:

Given that

At top $d_2=30 mm,P_2=860 KPa ,P_1=1000 KPa,d_1=85 mm$

$\rho =900\dfrac{Kg}{m^3}$

We know that

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2$

$A_1V_1=A_2V_2$

$\frac{V_1}{V_2}=\left(\dfrac{d_2}{d_1}\right)^2$

$\frac{V_1}{V_2}=\left(\dfrac{30}{85}\right)^2$

$V_2=8.02V_1$

$Z_2=12 sin60^{\circ}$

$\dfrac{1000\times 1000}{900\times 9.81}+\dfrac{V_1^2}{2\times 9.81}+0=\dfrac{860\times 1000}{900\times 9.81 }+\dfrac{V_2^2}{2\times 9.81}+12 sin60^{\circ}$

So $V_1=1.30$ m/s

We know that flow rate Q=AV

$Q=A_1V_1$

By putting the values

$A_1=\dfrac{\pi}{4}d^2$

$Q=7.3\times 10^{-3} m^3/s$

To find the flow rate we do not need the direction of flow,because we are just doing balancing of energy at inlet and at the exits of pipe.

4 0

3 years ago