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3 years ago

What is the electromagnetic spectrum used for?

Physics

2 answers:

KATRIN_1 [288]3 years ago

8 0

Answer:

The electromagnetic spectrum is used for identifying what type a wave is. For example, a wave with a very low frequency would be a radio wave, and a wave with a very high wavelength would be a gamma wave.

Let me know if this helps!

stiks02 [169]3 years ago

5 0

Answer: Astronomers use the entire electromagnetic spectrum to observe a variety of things. Radio waves and microwaves – the longest wavelengths and lowest energies of light – are used to peer inside dense interstellar clouds and track the motion of cold, dark gas

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Assume that the Styrofoam slab and the fur are both initially neutral, and that the slab charged negatively after it is rubbed w

BARSIC [14]

Charging phenomenon is basically transfer of electrons from one system to other. When a system loses the electrons the it becomes positively charged and when a system gains electrons then it will become negatively charged

so here if slab is rub against fur then one of them will lose the electrons while other will gain the electrons and hence one will get negatively charged and other will get positive charge

Here it is given that slab get negatively charged so correct options are

3. Electrons move from the fur to the slab.

6. The fur becomes positive after rubbing the slab.

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3 years ago

Un automovil con la velocidad de 5 m/s acelera durante 12 s a 3 m/s2,¿cual es la velocis

harina [27]

Answer:

60 miles

Explanation:

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3 years ago

Please help!!!! Will mark brainliest.

julia-pushkina [17]

Answer:

Approximately $8.4 \times 10^{2}\; \rm N$ , assuming that $g = 9.8\; \rm m \cdot s^{-2}$ .

Explanation:

Let $m$ and $a$ denote the mass and acceleration of Spiderman, respectively.

There are two forces on Spiderman:

Downward gravitational attraction from the earth: $W = m \cdot g$ .
Upward tension force from the strand of web $F(\text{tension})$ .

The directions of these two forces are exactly opposite of one another. Besides, because Spiderman is accelerating upwards, the magnitude of $F(\text{tension})$ (which points upwards) should be greater than that of $W$ (which points downwards towards the ground.)

Subtract the smaller force from the larger one to find the net force on Spiderman:

$(\text{Net Force}) = F(\text{tension}) - W$ .

On the other hand, apply Newton's Second Law of motion to find the value of the net force on Spiderman:

$(\text{Net Force}) = m \cdot a$ .

Combine these two equations to get:

$m \cdot a = (\text{Net Force}) = F(\text{tension}) - W$ .

Therefore:

$\begin{aligned}& F(\text{tension})\\ &= m \cdot a + W \\ &= m \cdot (a + g)\\ &= 76\; \rm kg \times \left(1.3\; \rm m \cdot s^{-2} + 9.8\; \rm m \cdot s^{-2}\right)\\ &\approx 8.4\times 10^{2}\; \rm N\end{aligned}$ .

By Newton's Third Law of motion, Spiderman would exert a force of the same size on the strand of web. Hence, the size of the force in the strand of the web should be approximately $8.4\times 10^{2}\; \rm N$ (downwards.)

4 0

3 years ago

by scale drawing, find the resultant of vectors 70N inclined at100N and direction of the resultant 100N

Thepotemich [5.8K]

Answer: 70n=

100n=

Explanation:

4 0

3 years ago

A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

vlabodo [156]

Answer:

r < a:

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

a < r < b:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

r = b:

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

b < r < c:

E = 0

r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

r < c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

r<a:

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

At r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

At a < r < b:

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

$E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = b:

$E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

Again, the minus sign indicates the direction of the field towards the center.

At b < r < c:

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

At r < c:

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

3 0

3 years ago