The density of an object is dependent upon the object’s mass and ---

When the bug is stationary and creating waves, how does the frequency of the wave some distance away from the bug compare with t

Brrunno [24]

Answer:

The frequency is the same

Explanation:

When a wave is created by a source which is vibrating at a certain frequency, the frequency of the wave itself is equal to the frequency of the source.

This occurs with every kind of wave. For instance, if we consider the radio waves produced by an antenna, the frequency of the radio waves is equal to the frequency of the antenna.

In this case, the waves are created by the vibrating bug. The bug is vibrating with a certain frequency $f$ : as a consequence, the frequency $f'$ of the waves produced by the bug will be equal to the frequency of vibration of the bug:

$f'=f$ .

3 0

2 years ago

Sound travels Faster through _____________ mediums.

Maru [420]

Answer:

I know for sure that it's either A or B have a great day sorry I don't know which one completely

3 0

2 years ago

Read 2 more answers

How r u?<br> Are u good?

Maslowich

Answer:

i'm good, hru hope ur staying safe

Explanation:

6 0

2 years ago

Read 2 more answers

You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla

lesantik [10]

Answer:

When they are connected in series

The 50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

The power rating of the first bulb is $P_1 = 100 \ W$

The power rating of the second bulb is $P_2 = 50 \ W$

Generally the power rating of the first bulb is mathematically represented as

$P_1 = V^2 R$

Where $V$ is the normal household voltage which is constant for both bulbs

So

$R_1 = \frac{V^2}{P_1 }$

substituting values

$R_1 = \frac{V^2}{100}$

Thus the resistance of the second bulb would be evaluated as

$R_2 = \frac{V^2}{50}$

From the above calculation we see that

$R_2 > R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 1 = I^2_1 R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 2 = I^2_2 R_2$

Now given that they are connected in series which implies that the same current flow through them so

$I_1^2 = I_2^2$

This means that

$P \ \alpha \ R$

So when they are connected in series

$P_2 > P_1$

This means that the 50 W bulb glows more than the 100 \ W bulb

3 0

3 years ago

DochEvi [55]

Answer:

Explanation:

A novae in astronomy means an explosion in the white dwarf star which had tapped enough gas from a companion star,hence it releases an incredible amount of energy which is Over a million times brighter than it normal stars.

A super novae on the other hand is a cosmic explosion that can be a billion times brighter than the normal.

From this one can see that a perculiar similarity between a novae and super novae is that both generate huge explosion and bright Ness, and a major difference is super novae release huge amount of brightness and energy more than the novae

3 0

2 years ago