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Evgen [1.6K]
3 years ago
9

Batteries typically have ratings stated in ampere-hours, that let you estimate the length of time any particular current level c

an be delivered before discharging. How much energy is approximately stored by a fully charged 50 A-hour, 12 V battery?

Engineering
1 answer:
DerKrebs [107]3 years ago
6 0

Answer: attached

Explanation:

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The in situ moisture content of a soil is 18% and the moist (total) unit weight is 105 pcf. The soil is to be excavated and tran
densk [106]

Answer: Volume= 1.16 yd³

Explanation:

We are given the in-situ moisture content= 18% and moist unit weight= 105 pcf

First we find out the dry unit weight

Dry unit weight= Moist unit weight / (1+ moisture content)

Dry unit weight= \frac{105}{1.18}

Dry unit weight= 88.983 pcf

Now, to find the volume ( in terms of each cubic yard i.e 1 yd³ we have

Volume = Dry unit weight compacted / Dry unit weight (in-situ) * 1 yd³

Volume= \frac{103.5}{88.983} * 1

Volume= 1.16 yd³

6 0
4 years ago
A parallel circuit with two branches and an 18 volt battery. Resistor #1 on the first branch has a value of 220 ohms and resisto
likoan [24]

Answer:

  2.455 W

Explanation:

The power dissipated in each branch is ...

  P = V^2/R

So, the branch powers are ...

  branch 1: 18^2/220 ≈ 1.473 W

  branch 2: 18^2/330 ≈ 0.982 W

Total power is ...

  1.473 W + 0.982 W = 2.455 W

8 0
3 years ago
Find the resultant of two forces 130 N and 110 N respectively, acting at an angle whose tangent
monitta

Answer:

F_r = 200N

Explanation:

Given

Let the two forces be

F_1 = 130N

F_2 = 110N

and

\tan(\theta) = \frac{12}{5}

Required

Determine the resultant force

Resultant force (Fr) is calculated using:

F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)

This means that we need to first calculate \cos(\theta)

Given that:

\tan(\theta) = \frac{12}{5}

In trigonometry:

\tan(\theta) = \frac{Opposite}{Adjacent}

By comparing the above formula to \tan(\theta) = \frac{12}{5}

Opposite = 12

Adjacent = 5

The hypotenuse is calculated as thus:

Hypotenuse^2 = Opposite^2 + Adjacent^2

Hypotenuse^2 = 12^2 + 5^2

Hypotenuse^2 = 144 + 25

Hypotenuse^2 = 169

Hypotenuse = \sqrt{169

Hypotenuse = 13

\cos(\theta) is then calculated using:

\cos(\theta)= \frac{Adjacent}{Hypotenuse}

\cos(\theta)= \frac{5}{13}

Substitute values for F_1, F_2 and cos(\theta) in

F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)

F_r^2 = 130^2 + 110^2 + 2*130*110*\frac{5}{13}

F_r^2 = 16900 + 12100 + 11000

F_r^2 = 40000

Take square roots of both sides

F_r = \sqrt{40000

F_r = 200N

<em>Hence, the resultant force is 200N</em>

4 0
3 years ago
Hello how are you you are loved
alexandr1967 [171]

Answer:

thx

Explanation:

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3 years ago
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Answer:

iejf go jjflkkkfffnfb ffv jedj HD video for my homework and I will be

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