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3 years ago

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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?

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

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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

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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>

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alexandr1967 [171]

Answer:

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Explanation:

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Answer:

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