To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as
Here,
Q = Total Heat
T = Temperature
The total change of entropy from a cold object to a hot object is given by the relationship,
From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'
Change in entropy is smaller than
Therefore the correct answer is C. Will always have a smaller magnitude than the change in entropy of the cold object
<h2>Answer</h2>
1m/s
<h2>Explanation</h2>
Given that:
<em>Mass of first blob = 2kg = m1</em>
<em>Velocity of blob = 4m/s = v1</em>
<em>Mass of second blob = 6kg = m2</em>
<em>Velocity of blob = 0m/s = v2</em>
<em />
To find:
<em>Final velocity = Vf</em>
<em />
<em>This question is of inelastic collision which is any collision between objects in which some energy is lost.</em>
<em />
<h3>Formula to be use:</h3><h2>(m1*v1) + (m2*V2) = Vf(m1 + m2)</h2>
(2*4) + (6*0) = Vf(2+6)
8 + 0 = Vf(8)
8 = Vf(8)
Vf = 1 m/s
So the speed of two blobs immediately after colliding = 1 m/s
Answer:
) the uniform disk has a lower moment of inertia and arrives first.
Explanation:
(a) the uniform disk has a lower moment of inertia and arrives first.
(b) Let's say the disk has mass m and radius r, and
the hoop has mass M and radius R.
disk: initial E = PE = mgh
I = ½mr², so KE = ½mv² + ½Iω² = ½mv² + ½(½mr²)(v/r)² = (3/4)mv² = mgh
m cancels, leaving v² = 4gh / 3
hoop: initial E = Mgh
I = MR², so KE = ½MV² + ½(MR²)(V/R)² = MV² = Mgh
M cancels, leaving V² = gh
Vdisk = √(4gh/3) > Vhoop = √(gh)
Answer:
9.2 m/s
Explanation:
v^2 = u^2 + 2as
u = 4.71
a= g * sin(8.30)
v= sqrt( 4.71^2 + 2 * 9.8* sin( 8.30 deg) * 22) = 9.2 m/s