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RideAnS [48]
3 years ago
14

Name two types of battery chargers that are used in mechanics

Engineering
1 answer:
tekilochka [14]3 years ago
7 0
Primary batteries
Secondary batteries
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A moving-coil instrument, which gives full-scale deflection with 0.015 A has a copper coil having resistance of 1.5 Ohm at 15°C
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Answer: check answers in pictures (2 p)

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3 years ago
What do you own that might not be manufactured?
horrorfan [7]

Answer:

A pet

Explanation:

Latin time I checked animals aren't made by people? I honestly don't know if this helps but I'm technically not wrong.

8 0
3 years ago
The 240-ft structure is used to provide various support services to launch vehicles prior to liftoff. In a test, a 12-ton weight
alina1380 [7]

Answer:

hello your question lacks the required question attached below is the missing diagram

Forces in GJ = -4.4444 i.e. 4.4444 tons

Forces in IG = 15.382 tons ( T )

Explanation:

Forces in GJ = -4.4444 i.e. 4.4444 tons

Forces in IG = 15.382 tons ( T )

attached below is the detailed solution

3 0
3 years ago
Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t
kodGreya [7K]

Answer:

\frac{T-20}{130-20}= e^{-14.28*0.05}

And if we solve for T we got:

T= 20 + 110e^{-14.28*0.05} = 73.86 C

The answer for this case would be T = 73.86 C at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

h = 20 \frac{W}{m^2 K} represent the heat transfer coefficient.

p represent the perimeter for this case and would be given by:

p = 2*0.05m +2*0.001m= 0.102m

k = 200 \frac{W}{m C} represent the thermal conductivity

w = 5cm =0.05 m represent the width

h = 1mm =0.001m represent the thickness

A= wh= 0.05m *0.001m = 0.00005 m^2

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}

Where m is a coefficient given by:

m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}

The value of x for this case represent the distance x =5 cm =0.05m

T_b =130 C represent the base temperature

T_{\infty}= 20 represent the temperature of the sorroundings or the ambient.

If we replace we have this:

\frac{T-20}{130-20}= e^{-14.28*0.05}

And if we solve for T we got:

T= 20 + 110e^{-14.28*0.05} = 73.86 C

The answer for this case would be T = 73.86 C at 5cm from the base of the fin.

3 0
3 years ago
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