Answer:
period of oscillations is 0.695 second
Explanation:
given data
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force
force = k × x .........1
so force = mg = 0.35 (9.8) = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π × ................2
put here value
period of oscillations = 2π ×
period of oscillations = 0.6953
so period of oscillations is 0.695 second
Answer:
Final temperature is 295K
Explanation:
Where the sample of copper is placed in the water, the heat transferred from the copper is equal that the heat absorbed by the water.
The heat transferred from the copper is:
C××mass×ΔT
Where C is molar heat capacity of copper (24,5J/molK)
Mass is 25,0g
And ΔT is final temperature - initial temperature (X-363K)
Also, the heat absorbed by the water is:
-C××mass×ΔT
Where C is molar heat capacity of water (75,2J/molK)
Mass is 100,0g
And ΔT is final temperature - initial temperature (X-293K)
As heat transferred is equal to heat absorbed:
24,5J/molK××25,0g×(X-363K) = -75,2J/molK×
×100,0g× (X-293K)
9,64X J/K - 3499J = - 417X J/K + 122273J
426,64X J/K = 125772 J
<em>X = 295K</em>
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Final temperature is 295K
I hope it helps!