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3 years ago

In 2-4 sentences, summarize what you know about EM waves.

Physics

2 answers:

MaRussiya [10]3 years ago

8 0

Answer:

In physics, electromagnetic radiation refers to the waves of the electromagnetic field, propagating through space, carrying electromagnetic radiant energy. It includes radio waves, microwaves, infrared, light, ultraviolet, X-rays, and gamma rays.

umka21 [38]3 years ago

8 0

I agree to what he said^

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Please answer this question with an explanation<br> will mark brainiest

irga5000 [103]

Answer:

jenny

Explanation:

6 0

3 years ago

Read 2 more answers

A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x

DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is $\lambda$

Total charge, Q = 7 nC = $7\times 10^{- 9} C$

At x = 2L,

Electric field, $\vec{E_{2L}} = 500N/C$

Coulomb constant, K = $8.99\times 10^{9} N.m^{2}/C^{2}$

Now, we know that:

$\vec{E} = K\frac{Q}{x^{2}}$

Also the line charge density:

$\lambda = \frac{Q}{L}$

Thus

Q = $\lambda L$

Now, for small element:

$d\vec{E} = K\frac{dq}{x^{2}}$

$d\vec{E} = K\frac{\lambda }{x^{2}}dx$

Integrating both the sides from x = L to x = 2L

$\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx$

$\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]$

$\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}$

Similarly,

For the field in between the range 2L< x < 3L:

$\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx$

$\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]$

$\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}$

Now,

If at x = 2L,

$\vec{E_{2L}} = 500 N/C$

Then at x = 3L:

$\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C$

4 0

4 years ago

What is the equation of universal law of garavitation

Vladimir79 [104]

Answer:

$F= \frac{GmM}{ {r}^{2} }$

G= gravitational constant

F= Gravitational force

m= mass of object 1

M= mass of object 2

r= distance between two objects

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3 years ago

In a nuclear power plant, the temperature of the water in the reactor is above 100°C because of what?

SVETLANKA909090 [29]

Answer:

The temperature of the water increases because the nuclear reactor heats it producing steam

Explanation:

The nuclear power plants are usually defined as those thermal plants where the nuclear reactors are used in order to generate heat that eventually leads to the rotating of the turbines and produces electricity. Here the nuclear reactor heats the water, and it increases above a temperature of 100°C, where this heat energy plays a key role in the entire process. It is an efficient method as it does not lead to the emission of any green house gases that are harmful to the environment.

8 0

3 years ago

Introduction: The specific heat capacity of a substance is the amount of energy needed to change the temperature of that substan

ki77a [65]

Answer:

A) 8,368 J

B) ) 0.893 J/gºC

Explanation:

The heat gained by the water can be obtained solving the following equation:

$q_{g} = c_{w} * m * \Delta T (1)$

where cw = specific heat of water = 4.184 J/gºC
m= mass of water = 1,000 g
ΔT = 2ºC
Replacing these values in (1) we get:

$q_{g} = c_{w} * m * \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)$

Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative: -8,368 J.
Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:

$q_{l} = c_{Al} * m_{Al} * \Delta T (3)$

⇒ $-8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)$

$c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)$

which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.

8 0

3 years ago