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3 years ago

8

Why is alternating current more effective at long–distance travel than direct current?

1 answer:

alina1380 [7]3 years ago

3 0

If you are using PLATO, which I'm sure you are cause I've had the same question, the answer it the following:

Transformers increase and decrease voltage as needed.

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How many kindergarteners do you think you can take on a fight before getting tired or over powered?

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Is this a serious question ?

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3 years ago

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calculate the work done by a girl of mass 40 kg when she climbs a stair case of 20 steps each of height 10 cm and acceleration i

Volgvan

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i think its a 800x² hope you like it

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3 years ago

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Dinah is playing cards with her friends, and it is her turn to deal. A card with a mass of 2.3 g slides 0.35 m along the table b

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=2.3+0.35-0.24

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2 years ago

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

The height of the cylinder is $h = 0.588\ m$

The face Area is $A = 4.19 \ m^2$

The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

The weight of the cylinder = Upward thrust

This is mathematically represented as

$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

4 0

3 years ago

How can a megaphone used by a cheerleader allow people on the opposite side of a field to hear her better?

Goshia [24]

The megaphone is cupping the area around her mouth and it is like with your hands, if you cover around them the noise will be louder. It is because it is hollow and there isn't anything to block the noise through the megaphone.

7 0

3 years ago