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jek_recluse [69]
3 years ago
5

If the frequency of an electromagnetic wave is too high or too low. A) It cannot be seen. B) It is no longer a wave. C) It is no

longer an electromagnetic wave. D) It becomes a sound wave
Physics
1 answer:
CaHeK987 [17]3 years ago
7 0
If the frequency of an electromagnetic wave is too high or too low : C. It is no longer an electromagnetic wave
If the frequency is too high or too low,  the skin depth pf the electromagnetic wave will not be comparable to the plasma length

hope this helps
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Un alumno menciona que al abrir la ventana de su casa sintió cómo el frío ingresaba a su cuerpo. Menciona cuál es la verdadera r
stepan [7]

Answer:

My believe the answer is

A.) or B.)

Explanation:

Here is why I think A is the answer.

If we use the process of elimination, it would look like this,

a) Porque el aire tiene una temperatura menor que la de su cuerpo; por eso se propaga más rápido.

<em>This makes sense because we all know in winter the weather is very cold and freezing.</em>

b) Porque la temperatura de su cuerpo, siente el aire frio que entra por la ventana.

<em>I feel like this answer is the question, but it could also be an answer, sorry, I'm a little uncertain.</em>

c) Porque el calor de su cuerpo se propaga al medio ambiente, al ser la temperatura del niño mayor que la del aire exterior.

<em>This answer has nothing to do with the question, plus it is very false, our body heat is not enough to overcome the very cold temperature from outside.</em>

d) Porque la temperatura del aire es igual a la temperatura del cuerpo.

<em>This is false because again our body heat is not even compared to the freezing cold temperatures from the winter.</em>

<em />

<em />

<h2>Well, have a nice rest of the day!</h2><h3>ba baiii!</h3>
3 0
3 years ago
What does the angular momentum quantum number determine? Check all that apply.
amm1812
I think the correct answers from the choices listed above are options 1, 5 and 7. Angular momentum quantum number determine the energy of an orbital, the shape of the orbital and <span>the overall size of an orbital. Hope this answers the question.</span>
8 0
4 years ago
Read 2 more answers
A ball is thrown straight up. What will
Papessa [141]

Answer:

velocity at the top: 0 m/s

acceleration at the top: -9.8 m/s²

Explanation:

Assuming up is positive and down is negative;

The velocity of the ball at the top of its path will be 0 m/s and the acceleration will be negative.

The velocity is 0 m/s because the ball does not move at the top of its path, and it switches from a positive velocity to a negative velocity. It must go through 0 in order to go from positive to negative.

The acceleration, however, is always negative no matter where the ball is in its motion. This negative acceleration causes the ball to slow down as it reaches the top, and speed up as it reaches the bottom.

<u>Think about it:</u> If there wasn't a negative acceleration, and it was instead 0, the ball would never come back down and instead keep going in a straight line.

6 0
2 years ago
Read 2 more answers
PLEASE ITS AN Emergency IF ITS RIGHT I WILL GIVE BRAINLIEST
n200080 [17]

Answer:

all of those are pisitions

Explanation:

6 0
3 years ago
An ideal spring hangs from the ceiling. A 1.95 kg mass is hung from the spring, stretching the spring a distance d=0.0865 m from
uranmaximum [27]

Answer:

kinetic energy = 0.1168 J

Explanation:

From Hooke's law, we know that ;

F = kx

k = F/x

We are given ;

Mass; m = 1.95 kg

Spring stretch; d = x = 0.0865

So, Force = mg = 1.95 × 9.81

k = 1.95 × 9.81/0.0865 = 221.15 N/m

Now, initial energy is;

E1 = mgL + ½k(x - L)²

Also, final energy; E2 = ½kx² + ½mv²

From conservation of energy, E1 = E2

Thus;

mgL + ½k(x - L)² = ½kx² + ½mv²

Making the kinetic energy ½mv² the subject, we have;

½mv² = mgL + ½k(x - L)² - ½kx²

We are given L=0.0325 m

Plugging other relevant values, we have ;

½mv² = (1.95 × 9.81 × 0.0325) + (½ × 221.15(0.0865 - 0.0325)² - ½(221.15 × 0.0865²)

½mv² = 0.62170875 + 0.3224367 - 0.82734979375

½mv² = 0.1168 J

7 0
3 years ago
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