What is the relationship between electric and magnetic fields?

Water enters a shower head

zalisa [80]

Answer:

13.1

Explanation:

thats what i put in for acellus and its right

4 0

2 years ago

Heat is transferred from the sun to the earth via electromagnetic waves (see Chapter 24). Because of this transfer, the entropy

Sever21 [200]

Answer:

1. Decreases,

2. Increases,

3. Increases

Explanation:

The heat which is a product of sun's energy, is transferred from the sun to the earth through radiation, conduction or convention. This heat passes through the earth atmosphere, then warms it , before becoming heat energy.

Therefore, Heat is transferred from the sun to the earth via electromagnetic waves . Because of this transfer, the entropy of the sun DECREASES, the entropy of the earth INCREASES and the entropy of the sun-earth system INCREASES.

8 0

3 years ago

Valence electrons determine an atom’s __________. a. mass c. atomic number b. chemical properties d. period Please select the be

Lemur [1.5K]

Hello,

Here is your answer:

The proper answer to this question is option B "<span>chemical properties".

Here is how:

</span>Electrons determine a atoms chemical properties mainly what its made of.

Your answer is B.

If you need anymore help feel free to ask me!

Hope this helps!

5 0

3 years ago

Read 2 more answers

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

3 years ago

The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

$\Delta x = \Delta y$

Where:

$\Delta x$ - Range of the bullet, measured in meters.

$\Delta y$ - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

$K_{1} = U_{2}$