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PtichkaEL [24]
2 years ago
14

Objects A and B both have mass 2 kg. Object A has temperature 20°C and Object B has temperature 40°C. The specific heat of Obj

ect A is larger than that of Object B. The two objects are isolated from the environment and are brought into thermal contact with each other and allowed to come to thermal equilibrium. Is the final temperature of both objects greater than, less than or equal to 30°C? Briefly explain your reasoning.
Physics
1 answer:
OverLord2011 [107]2 years ago
8 0

Answer:

Equal to 30°C.

Explanation:

The final temperature of both objects are equal to 30°C because of the conduction heat from one object to another. Both objects have same mass of 2 kg and both were isolated from the external environment so conduction of heat occurs from the more hotter body to the less hotter body until the temperature of both objects are equal to each other.

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The pressure in Arabian sea is much bigger than near the surface?<br> please don't scam
sashaice [31]

Answer:

In general, the annual sea surface temperatures(SSTs) in the Bay of Bengal(BOB) are higher than the Arabian sea(AS). because, there are two main reasons for higher SST in the Bay of Bengal than the Arabian Sea. they are 1. stratification, 2.strong mixing

stratification is nothing but a phenomenon which stratifies(layers) the sea water when different density water(fresh water, rain water) add into the sea water. So the stratification in the bay of Bengal is comparatively high than the Arabian sea due to the high river discharge and precipitation in the BOB than the AS. the mixing process over the Arabian sea is higher than the Bay of Bengal due to the prevailing of strong winds over the AS (because of the presence of the mountains of east Africa) than Bay of Bengal (because of the winds over the BOB are sluggish in nature then the AS). But generally winds over the sea mixes easily the normal sea water than stratified/stabilized sea water column. That's why less mixing will takes place over the surface of BOB than the AS. So due to the presence of less mixing over the surface of the Bay of Bengal than the Arabian sea, the SST values over the Arabian sea are always lower than the Bay of Bengal. that's why the Arabian sea is colder than the Bay of Bengal.

Explanation:

4 0
2 years ago
Read 2 more answers
Brownian motion is due to:
My name is Ann [436]

Answer:

b

Explanation:

Brownian motion is the random movement of particles in a fluid due to their collisions with other atoms or molecules. ... Brownian motion takes its name from the Scottish botanist Robert Brown, who observed pollen grains moving randomly in water. He described the motion in 1827 but was unable to explain it.

8 0
3 years ago
A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t
postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water m=1.4\ kg

at 200^{\circ} specific volume is \nu =0.001157\ m^3\kg(From Table A-4,Saturated water Temperature table)

V_1=m\nu _1

V_1=1.4\times 0.001157

V_1=1.6198\times 10^{-3}\ m^3

Final Volume V_2=4V_1

V_2=4\times (1.6198\times 10^{-3})

V_2=6.4792\times 10^{-3}\ m^3

Specific volume at this stage

\nu _2=\frac{V_2}{m}

\nu _2=\frac{6.4792\times 10^{-3}}{1.4}

\nu _2=0.004628\ m^3/kg

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})

T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)

T_2=370.7^{\circ} C

4 0
3 years ago
As the wavelength of A wave in a uniform medium increases, its speed will what?
viva [34]

The speed of a wave in a uniform medium doesn't depend on its wavelength.


8 0
2 years ago
A closed-end organ pipe is used to produce a mixture of sounds. The third and fifth harmonics in the mixture have frequencies of
nexus9112 [7]

Answer:

F_1=366.67Hz

Explanation:

From the question we are told that:

Frequency of 3rd harmonics F_3=1100

Frequency of 5th harmonics F_3=1833

Generally the equation for Wavelength at 3rd Harmonics is mathematically given by

 \lambda_3=\frac{4}{3}l

Therefore

 F_3=\frac{3v}{4l}

Generally the equation for Wavelength at 1st Harmonics is mathematically given by

 \lambda_1=\frac{4}{1}l

Therefore

 F_1=\frac{v}{4l}

Generally the equation for the frequency of the first harmonic is mathematically given by

 F_1=\frac{F_3}{3}

 F_1=\frac{1100}{3}

 F_1=366.67Hz

7 0
2 years ago
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