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3 years ago

To make an even better electrical junction, what should you do?

Engineering

1 answer:

docker41 [41]3 years ago

4 0

Answer:

A:Solder it.

Explanation:

Hopefully this helps!

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Why is sitting correctly important

castortr0y [4]

Answer: It helps your back posture and shows respect

Explanation:

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3 years ago

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Nancy ate a 500 Cal lunch. Neglecting efficiency issues (i.e., assuming 100% conversion of energy to work), to what height could

VladimirAG [237]

Answer:

$4265.04\ \text{m}$

$2.38\times 10^{10}\ \text{W}$

Explanation:

PE = Energy of food = 500 cal = $500\times4184=2.092\times10^6\ \text{J}$

m = Mass of object = 50 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Potential energy of food is given by

$PE=mgh\\\Rightarrow h=\dfrac{PE}{mg}\\\Rightarrow h=\dfrac{2.092\times 10^6}{50\times 9.81}\\\Rightarrow h=4265.04\ \text{m}$

Nancy could raise the weight to a maximum height of $4265.04\ \text{m}$ .

Mass of $H_2$ used per year = $25\times 10^{9}\ \text{kg/year}$

Energy of $H_2$ = $\dfrac{30\times10^9}{1000}=30\times 10^6\ \text{J/kg}$

Power

$P=25\times 10^{9}\ \text{kg/year}\times 30\times 10^6\ \text{J/kg}\\\Rightarrow P=7.5\times 10^{17}\ \text{J/year}\\\Rightarrow P=\dfrac{7.5\times 10^{17}}{365.25\times 24\times 60\times 60}\\\Rightarrow P=2.38\times 10^{10}\ \text{W}$

The power requirement is $2.38\times 10^{10}\ \text{W}$ .

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3 years ago

18. What is being shown in the above Figure?

slavikrds [6]

D. Camshaft gear backlash is being checked

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3 years ago

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14. Tires are rotated to

ExtremeBDS [4]

Answer:

Explanation:

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3 years ago

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Calculate the amount of power (in Watts) required to move an object weighing 762 N from point A to point B within 29 seconds. Di

Tresset [83]

Answer:

0.556 Watts

Explanation:

w = Weight of object = 762 N

s = Distance = 5 m

t = Time taken = 29 seconds

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow a=\frac{2\times (s-ut)}{t^2}\\\Rightarrow a=\frac{2\times (5-0)}{29^2}=\frac{10}{481}$

Mass of the body

$m=\frac{w}{g}=\frac{762}{9.81}$

Force required to move the body

$F=ma\\\Righarrow F=\frac{762}{9.81}\times \frac{10}{481}$

Velocity of object

$v=u+at\\\Rightarrow v=0+\frac{10}{481}\times 29\\\Rightarrow v=\frac{10}{29}$

Power

$P=Fv\\\Rightarrow P=\frac{762}{9.81}\times \frac{10}{481}\times \frac{10}{29}=0.556\ W$

∴ Amount of power required to move the object is 0.556 Watts

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4 years ago