A 0.20 kg baseball is traveling at 40 m/s toward the batter. The ball is hit by the bat with a force of 200N, and is
1 answer:
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Answer:
<em>1.</em> <em>39068.07 N</em>
<em>2. 19534.036 N</em>
Explanation:
depth of water h = 10 m
atmospheric pressure Patm = 101325 Pa
density of water p = 1000 kg/m^3
acceleration due to gravity g = 9.81 m/s^2
pressure due to depth of water = pgh
P = 1000 x 9.81 x 10 = 98100 Pa
total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa
Left plug has diameter = 50 cm = 0.5 m
radius = 0.5/2 = 0.25 m
height = 1 cm = 0.01 m
<em>height below tank surface = 10 - 0.01 = 9.99</em>
pressure at this depth = 1000 x 9.81 x 9.99 = 98001.9 Pa
<em>total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa</em>
surface area of plug = π = 3.142 x
= 0.196
<em>force required to lift left plug = pressure x area</em>
F = 199326.9 x 0.196 = <em>39068.07 N</em>
<em>The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug</em>
A = 0.196/2 = 0.098
force F required to lift right plug = 199326.9 x 0.098 =<em> 19534.036 N</em>
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