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2 years ago

Which electrical device makes it possible to transmit electrical energy efficiently from a power plant to users?

Physics

2 answers:

Debora [2.8K]2 years ago

6 0

Solar pannels are the energy sorce that lets you transfer energy from a plant source to the use or the consumer

Aneli [31]2 years ago

3 0

Hey, Name's Jessy. I hope, I answer your question.

A Transformer. It current carrying wire.

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Electromagnetic induction means that moving a magnet through a loop of wire creates an electric current.

MAVERICK [17]

False. An Electromagnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field.

4 0

3 years ago

Read 2 more answers

Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o

VARVARA [1.3K]

The rate of heat loss by radiation is equal to -207.5kW

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

$5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }$

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so, we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

$K=Celsius+273.15$

So, converting we have:

$T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K$

Therefore, substituting the given information and calculating, we have:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

$HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW$

Hence, we have that the rate of heat loss is equal to -207.5kW.

8 0

2 years ago

A raft with the area A , thickness= h and the mass 600 kg, Floats in still water with 7 cm

elena55 [62]

In this problem, we need to solve for Bubba’s mass. To do this, we let A be the area of the raft and set the weight of the displaced fluid with the raft alone as ρwAd1g and ρwAd2g with the person on the raft, where ρw is the density of water, d1 = 7cm, and d2= 8.4 cm. Set the weight of displaced fluid equal to the weight of the floating objects to eliminate A and ρw then solve for m.

ρwAd1g = Mg

ρwAd2g = (M + m) g

d2∕d1 = (M + m)/g

m = [(d2∕d1)-1] M = [(8.4 cm/7.0 cm) - 1] (600 kg) =120 kg

This means that Bubba’s mass is 120 kg.

7 0

3 years ago

A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate

GaryK [48]

Answer:

It changes at a rate of 4/3 meter per second

Explanation:

In the given figure below we have

$\Delta OBD\simeq \Delta ABC\\\\\therefore \frac{5}{X+Y}=\frac{2}{Y}\\\\$

Solving for Y given $X=2m/s$ we get

$\frac{5}{2+Y}=\frac{2}{Y}\\\\5Y=4+2Y\\\\Y=\frac{4}{3}m/s$

8 0

2 years ago

beks73 [17]

Answer:

2 m/s²

Explanation:

the equations of motion are

S= ut +½at²

v² = u²+ 2as

v = u + at

s = (u+v)/2 × t

From the parameters given

u = 0m/s this is because it starts from rest

Distance (s) = 9m

Time (t) = 3s

Based on this the first equation would be used

s = ut + ½at²

Input values

9 = 0×3 + ½ × a x 3²

9 = 0 + 9a/2

9 = 4.5a

Divide both sides by 4.5

a = 9 / 4.5 m/s²

a = 2 m/s²

I hope this was helpful, please mark as brainliest

3 0

3 years ago