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Scorpion4ik [409]
2 years ago
8

Help ASAP Just answer the first question for me please!

Physics
1 answer:
Usimov [2.4K]2 years ago
4 0

Answer:

Im pretty sure its b

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You hear a sound with a frequency of 256 Hz. The amplitude of the sound increases and decreases periodically: it takes 2 seconds
german

To solve this problem it is necessary to take into account the concepts related to frequency and period, and how they are related to each other.

The relationship that defines both agreements is given by the equation,

f_{beat}=\frac{1}{T}

Then the frequency for the previous period given (2sec) is

f_{beat}=\frac{1}{2}

f_{beat} = 0.5Hz

The beat frequency of two frequencies is equal to the difference between the two frequencies, then

f_{beat} = |f_1-f_2|\\f_{beat} = |256Hz-2Hz|\\f_{beat} = 254Hz

<em>Hence option A is incorrect.</em>

We can do this process for 254Hz as f_1 and 258 Hz for f_2 , then

f_{beat} =|254Hz-258Hz|

f_{beat} = 4Hz

<em>Hence option B is incorrect. </em>

We can also do this process for 255Hz as  f_1 and 257 Hz for f_2 , then

f_{beat} =|255Hz-257Hz|

f_{beat} = 2Hz

<em>Hence option C is incorrect. </em>

We can also do this process for 255.5Hz as f_1 and 256.5 Hz for f_2, then

f_{beat} =|255.5Hz-256.5Hz|\\f_{beat} = 1Hz

<em>Hence option D is incorrect. </em>

We can also do this process for 255.75Hz as f_1 and 256.25 Hz for f_2, then

f_{beat} =|255.75Hz-256.25Hz|\\f_{beat} = 0.5Hz

<em>Hence option E is incorrect. </em>

Therefore the sum of the frequencies in the sound wave would be 256.25Hz and 255.75Hz

3 0
3 years ago
A summary of a source is usually
Mekhanik [1.2K]

Answer:

A shorter than the original source and in the researcher's words

Explanation:

The summary is an abridged version of the original source and in the researcher's own very words.

Summaries gives an over-arching perspective and excludes implicit details from a given text.

A summary should not be detailed and must avoid overt illustrations. They must capture the true essence of piece leaving out flowery details.

A summary should be lesser in length than the original piece. Any third party reader should immediately be able to grab the details of the original piece from the summary.

5 0
3 years ago
Read 2 more answers
A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri
GREYUIT [131]

Answer:

|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}

Now, substitute this into 'dF':

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)

Now we can integrate dF over the rod.

\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)

4 0
3 years ago
If the mass of a material is 47 grams and the volume of the material is 15 cm^3, what would the density of the material be?
uranmaximum [27]
Density is calculated as mass per unit volume. In this case, since the material has a mass of 47 grams and we have the volume of 15 cm^3, we can simply divide the values:
Density = 47 grams / 15 cm^3 = 3.1 g/cm^3
Therefore, the material has a density of 3.1 g/cm^3
8 0
3 years ago
Please answer all of these questions for brainly!
Nana76 [90]
Answer:

1. C
2. B
3. D
4. A
3 0
3 years ago
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