The pair of blocks is insulated, so no energy escapes. The pair of temperatures possible is 95 +95 temperature blocks.
<h3>What is thermal equilibrium?</h3>
When two objects are in direct contact and transfer heat through conduction. When the both object attain same temperature after sometime, they are called in thermal equilibrium.
Two identical blocks are heated to different temperatures. The blocks are placed so that they touch, and heat begins to flow between blocks. The heat will continue to until and unless they have same temperatures. After they being isolated, the temperature of both will be same and no heat is transferred outside.
Thus, the pair of temperatures possible is 95 +95 temperature blocks.
Learn more about temperature.
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Answer:


Explanation:
what is the smallest crater that each of these telescopes could resolve on our moon?
For moon ;
s = 3.8 × 10 ⁸ m
y = 1.22 λs/D
where;
λ = 400 nm = 400× 10 ⁻⁹
D = 2.4 m
The smallest crater for the hubble space is calculated as follows:


For Aceribo ;
y = 1.22 λs/D
where :
λ = 75 cm = 0.75 m
D = 305 m


Answer:
Person B
Explanation:
Person B will hear the frequency that is lower than the B- flat.
The volume would be 287cm³. Multiply all the 3 numbers by each other
Answer:
The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
Explanation:
Given that,
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.
The speed of sound in air is 343 m/s.
To find,
The wavelength range for the corresponding frequency.
Solution,
The speed of sound is given by the following relation as :

Wavelength for f = 45 Hz is,


Wavelength for f = 375 Hz is,


So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.