Answer:
0.63m
Explanation:
Volume expansivity = change in volume/original volume×temp change
Volume expansivity p = 3x
p = ∆V/V∆t
x is the linear expansivity
Given
x = 6 x 10^-4
p = 3x
p = 3(6 x 10^-5)
p = 18×10^-5
Volume = 100m³
∆t = 45-10 = 35°C
Required
Change in volume ∆V
Substitute the given values into the formula
18×10^-5 = ∆V/100(35)
18×10^-5 =∆V/3500
∆V = 3500×0.00018
∆V = 0.63m
Hence the increase in volume of the Mercury is 0.63n
Hi lamy
we know ΔU=qΔV
solving for potential difference ΔV
we get ΔV=ΔU/q
ΔV= (Ub-Ua)/q
Ub is potential energy at b
Ua is potential energy at a
q is the charge of the particle
plug in and find out
Answer:
Low-temperature blackbody
Explanation:
There are 3 types of blackbody temperatures.
Low-temperature blackbody
High temperature extended area blackbody
High-temperature cavity blackbody
A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.
The concept required to solve this problem is that related to the Isobaric process. Isobaric process is understood as the process in which changes occur at constant pressure. From the first law of thermodynamics this can be expressed as,
Here,
P = Pressure
dV = Differential of Volume
As the Pressure is constant we have,
Replacing
Therefore the correct answer is A.