Answer:
193.4 J/K
Explanation:
For a closed system, the entropy is given as the natural logarithm of microstates.
The number of particles ( grains of sand ) N = 28
The number of boxes ( compartments ) M = 1000
Entropy is the logarithm of number of microstates
But the number of microstates is given by
Answer:
option (D)
Explanation:
Here initial rotation speed is given, final rotation speed is given and asking for time.
If we use
A) θ=θ0+ω0t+(1/2)αt2
For this equation, we don't have any information about the value of angular displacement and angular acceleration, so it is not useful.
B) ω=ω0+αt
For this equation, we don't have any information about angular acceleration, so it is not useful.
C) ω2=ω02+2α(θ−θ0)
In this equation, time is not included, so it is not useful.
D) So, more information is needed.
Thus, option (D) is true.
Answer:
Explanation:
I'm not sure you can do this without just a bit more information. I can tell you what the mass of the water is when the rocks are removed. When we know that, we know the volume of the water that was displaced. whether or not this is enough information to determine the volume of the box, I'm not sure.
400 grams raises the box 1 cm.
The density of water = 1 gm / cm^3
400 grams of water = 400 mL or 400 cm^3
The volume of the displaced water = 400 cm^3
The volume a slice from the square prism is B*h
B = 400 cm^2
h = 1 cm
If the base is 400 cm^2 then each side is
s^2 = 400
sqrt(s^2)= sqrt(400)
s = 20
The volume of the box is 20^3 = 8000 cm^3
Answer:
final temperature of slug of gold is 37°C
final temperature of slug of manganese is 28.56°C
Explanation:
Hello! To solve this problem we must take into account the concept of heat capacity, this is defined as the ratio between energy and temperature rise.
In other words it is the amount of energy that is required to increase a temperature degree.
Taking into account the above we infer the following equation
where
C=heat capacity
(a) Differentiate the position vector to get the velocity vector:
<em>r</em><em>(t)</em> = (3.00 m/s) <em>t</em> <em>i</em> - (4.00 m/s²) <em>t</em>² <em>j</em> + (2.00 m) <em>k</em>
<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em> = (3.00 m/s) <em>i</em> - (8.00 m/s²) <em>t</em> <em>j</em>
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(b) The velocity at <em>t</em> = 2.00 s is
<em>v</em> (2.00 s) = (3.00 m/s) <em>i</em> - (16.0 m/s) <em>j</em>
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(c) Compute the electron's position at <em>t</em> = 2.00 s:
<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>
The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:
||<em>r</em> (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m
(d) In the <em>x</em>-<em>y</em> plane, the velocity vector at <em>t</em> = 2.00 s makes an angle <em>θ</em> with the positive <em>x</em>-axis such that
tan(<em>θ</em>) = (-16.0 m/s) / (3.00 m/s) ==> <em>θ</em> ≈ -79.4º
or an angle of about 360º + <em>θ</em> ≈ 281º in the counter-clockwise direction.