The SI unit of measurement for mass is the

What is occurring when a light wave goes through a pane of glass in a window?

ycow [4]

Answer:

transmission: the passing of a wave through an object

Explanation:

5 0

3 years ago

The amplitude of a paricular wave is 4.0 m. The crest to trough distance

kozerog [31]

Answer:

The crest to trough distance = 8 m

Explanation:

Given that,

The amplitude of a particular wave is 4.0 m.

We need to find the crest to trough distance.

We know that,

Amplitude = The distance from the base line to the crest or the the distance from the baseline to the trough.

It means,

Distance from crest to trough = 2(Amplitude)

= 2(4)

= 8 m

Hence, the crest to trough distance is equal to 8 m.

6 0

3 years ago

By what factor must the amplitude of a sound wave be decreased in order to decrease the intensity by a factor of 3?

s344n2d4d5 [400]

Answer:

Amplitude is decreased by a factor of $\sqrt3$ if intensity is decreased by a factor of 3.

Explanation:

Intensity of a sound wave is directly proportional to the square of its amplitude.

Therefore, if intensity is $I$ and amplitude is $A$ , then

$I=kA^2$ , where, $k$ is constant of proportionality.

Now, if intensity of sound wave is decreased by a factor of 3. So,

New intensity is, $I_{new}=\frac{I}{3}$

$I_{new}=kA_{new}^2\\\frac{I}{3}=kA_{new}^2$

Plug in $kA^2$ for $I$ . This gives,

$\frac{kA^2}{3}=kA_{new}^2\\A_{new}^2=\frac{A^2}{3}\\A_{new}=\sqrt{\frac{A^2}{3}}=\frac{A}{\sqrt{3}}$

Therefore, amplitude is decreased by a factor of $\sqrt3$ .

4 0

3 years ago

What is the mass of the object if it has a density of 657 g/mL and a volume of 32 mL? Show work!

Luden [163]

Answer:

The answer is 21024g/mL

Explanation:

Multiply 657 by 32:

657

× 32

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

21024

↳ 21024 g/mL

3 0

3 years ago

As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of

kykrilka [37]

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution decrease and the vapor pressure of the solution decrease .

Explanation:

Depression in freezing point :

$\Delta T_f=K_f\times m$

where,

$\Delta T_f$ =depression in freezing point =

$K_f$ = freezing point constant

m = molality ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

If molality of the solution in high the depression in freezing point of the solution will be more.
If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

$\frac{p_o-p_s}{p_o}=\chi_{solute}$

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of solution

$\chi_{solute}$ = Mole fraction of solute

$p_s\propto \frac{1}{\chi_{solute}}$

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.

8 0

3 years ago