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Korvikt [17]
3 years ago
14

A student allows the evaporating dish and the anhydrous salt for a long time on the lab bench. What will happen to the mass of t

he content of the evaporating dish as it cools and how could this be prevented? G
Physics
1 answer:
Olenka [21]3 years ago
4 0

Anhydrous salts absorb moisture from their surroundings when it gets cooled, hence the mass of the content would increase. It can be prevented if we allow the evaporating dish to cool in a dessicator. A dessicator is a glass or plastic container in which chemicals can be stored and allowed to cool. This container can be sealed. Dessicator would provide an environment free of moisture instead of keeping in atmospheric air having moisture content.


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The rate of flow of electric CHARGE past any point is described in the unit of electric CURRENT ... the Ampere.
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An ideal monatomic gas initially has a temperature of T and a pressure of p. It is to expand from volume V1 to volume V2. If the
yawa3891 [41]

Answer:

Isothermal :   P2 = ( P1V1 / V2 ) ,  work-done pdv = nRT * In( \frac{V2}{v1} )

Adiabatic : : P2 = \frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }  , work-done =

W = (3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)

Explanation:

initial temperature : T

Pressure : P

initial volume : V1

Final volume : V2

A) If expansion was isothermal calculate final pressure and work-done

we use the gas laws

= PIVI = P2V2

Hence : P2 = ( P1V1 / V2 )

work-done :

pdv = nRT * In( \frac{V2}{v1} )

B) If the expansion was Adiabatic show the Final pressure and work-done

final pressure

P1V1^y = P2V2^y

where y = 5/3

hence : P2 = \frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }

Work-done

W = (3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)

Where    T2 = T1V1^(2/3)/V2^(2/3)

3 0
3 years ago
Using Kepler's 3rd law and Newton's law of universal gravitation, find the period of revolution P of the planet as it moves arou
weeeeeb [17]

Answer:

P =  \pi \sqrt{\frac{(R_1+R_2)^3}{2 \ GMS}}

Explanation:

From the attached diagram below:

AC = a (1 + e) = R₂     -------- equation (1)

CD = a ( 1 - e) = R₁     ---------   equation (2)

⇒ 1 - e = \frac{R_1}{a}

e= 1 - \frac{R_1}{a}

Replacing the value for e into equation (1)

a(1+1- \frac{R_1}{a})= R_2

= 2a - R_1 = R_2

a= \frac{R_1+R_2}{2}

From Kepler's third law;

P = 2 \pi \sqrt{\frac{a^3}{GMS}}

P = 2 \pi \sqrt{\frac{(R_1+R_2)^3}{8GMS}}

P =  \pi \sqrt{\frac{(R_1+R_2)^3}{2 \ GMS}}

5 0
3 years ago
Samples of different materials, A and B, have the same mass, but the sample
Effectus [21]

Answer:

B. The particles that make up material B have more mass than the

particles that make up material A.

Explanation:

3 0
3 years ago
The sampling rate of an ADC is 8.1 kHz. What will be an appropriate cut-off frequency (break frequency) for the anti-aliasing fi
KiRa [710]

Answer:

4000 Hz

Explanation:

An anti-alias filter is usually added in front of the ADC to limit a certain range of input frequencies in order to avoid aliasing. This filter is usually a low pass filter that passes low frequencies but attenuates the high frequencies.

The Nyquist sampling criteria states that the sampling rate should be at least twice the maximum frequency component of the desired signal.

Sampling rate = 2(max input frequency)

From the relation we can find out the cut-off frequency for the anti-aliasing filter.

max input frequency = sampling rate/2

max input frequency = 8100/2 = 4050 Hz

Therefore, 4000 Hz would be an appropriate cut-off frequency for the anti-aliasing filter.

3 0
3 years ago
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