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2 years ago

6

How much power do a capacitor and inductor dissipate? Assume the capacitor/inductor have no parasitic resistance (no resistor in

series with them), that is, they are ideal. If a voltage and current are delivered to the capacitor and inductor what is done with the energy they receive? Do the inductor or capacitor heat up when receiving energy? Why or why not? (4 points)

1 answer:

Mariana [72]2 years ago

4 0

They do in fact heat up while receiving energy.

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If particleboard is used as wall sheathing, the grade mark with type _____ or _____ should be stamped on it.

kramer

If particleboard is used as wall sheathing, the grade mark with type M1 or M2 should be stamped on it.

<h3>What is particle board?</h3>

Particle board is notably used as floors underlayment or as a base for parquet floors, timber floors, or for carpets. For this purpose, the particle forums are dealt with with unique chemical compounds and resins to cause them to water-resistant or termite proof.

Waferboard, OSB, and composite plywood, while carried out as wall sheathing, offer a nail base for software of shingle siding.

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2 years ago

What’s the number of gold atoms in a nanogram? a picogram?

zvonat [6]

Answer :

The number of gold atoms in nanogram is, $3.057\times 10^{12}$

The number of gold atoms in picogram is, $3.057\times 10^{9}$

Explanation :

As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

As we know that,

1 mole contains $6.022\times 10^{23}$ number of atoms.

First we have to determine the number of gold atoms in a nanogram.

As, 196.97 grams of gold contains $6.022\times 10^{23}$ number of gold atoms

And, 1 grams of gold contains $\frac{1g}{196.97g}\times (6.022\times 10^{23})$ number of gold atoms

So, $10^{-9}$ nanograms of gold contains $\frac{1g}{196.97g}\times (10^{-9})\times (6.022\times 10^{23})=3.057\times 10^{12}$ number of gold atoms

The number of gold atoms in nanogram is, $3.057\times 10^{12}$

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains $6.022\times 10^{23}$ number of gold atoms

And, 1 grams of gold contains $\frac{1g}{196.97g}\times (6.022\times 10^{23})$ number of gold atoms

So, $10^{-12}$ picograms of gold contains $\frac{1g}{196.97g}\times (10^{-12})\times (6.022\times 10^{23})=3.057\times 10^{9}$ number of gold atoms

The number of gold atoms in picogram is, $3.057\times 10^{9}$

8 0

3 years ago

Which of the following parasites is also known as ascarids? Roundworms hookworms whipworms or ear mites?

Jet001 [13]

Answer: Roundworm, hookworms, whipworms.

Explanation: Ascarids can be defined as any leaf insect or walking stick insect; phasmid nematode of the family Ascarididae.

They cause the disease ascariasis in humans and mammals and examples of them are roundworm, hookworm and whipworm.

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3 years ago

A vertical curve is designed for 55 mi h and has an intial grade of +2.5

Dafna1 [17]

Vertical curves are the curves which provide a vehicle to negotiate elevation rate change at a slow rate.

<u>Explanation:</u>

A vertical curve gives a progress between two slanted roadways, permitting a vehicle to arrange the height rate change at a slow rate as opposed to a sharp cut.

In any event four distinct criteria for building up lengths of list vertical bends are perceived somewhat. These are (1) front light sight separation, (2) rider comfort, (3) waste control, and (4) a dependable guideline for outward presentation.

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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