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zhannawk [14.2K]
3 years ago
10

What is the entropy of a closed system in which 28 distinguishable grains of sand are distributed among 1000 distinguishable equ

al-sized compartments?
Physics
1 answer:
jeyben [28]3 years ago
7 0

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

S=ln \psi

But the number of microstates is given by

\psi=M^{N} hence S=ln M^{N}=\ln (1000^{28})= \boxed{193.4 J/K}

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The weight of a body of certain mass becomes zero in space.why?write with reasons​
Anvisha [2.4K]

Answer:

Weight is what you get when a certain amount of gravity is acting on that mass, and something, like the surface of a planet, is resisting that action. In space, when falling freely, there's nothing resisting the pull of gravity so weight disappears. Mass however stays.

hope this helps u

Explanation:

7 0
3 years ago
In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s
sladkih [1.3K]

Answer:

  • quality factor (Q) = 69.99
  • inductor = 1.591 x 10⁻⁴ H
  • capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R =  10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW)  = \frac{R}{2\pi L}

BW = \frac{R}{2\pi L}

make L (inductor) the subject of the formula

L = \frac{R}{2\pi *BW}  =  \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH

F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}

make C (capacitor)  the subject of the formula

C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F

quality factor (Q) = \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99

quality factor (Q) =  69.99

5 0
3 years ago
Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
Coders play an important role in
Burka [1]

Answer:

it is a.health record documentation

Explanation:hope this helps

3 0
3 years ago
a motorcycle is capable of accelerating at 5.1 m/s starting from rest how far can it travell in 1.5 sec
Katarina [22]
The answer is 21m because the motion is in one dimension with constant acceleration.

The initial velocity is 0, because it started from rest, the acceleration <span>ax</span> is <span>4.7<span>m<span>s2</span></span></span>, and the time t is <span>3.0s</span>

Plugging in our known values, we have

<span>Δx=<span>(0)</span><span>(3.0s)</span>+<span>12</span><span>(4.7<span>m<span>s2</span></span>)</span><span><span>(3.0s)</span>2</span>=<span>21<span>m</span></span></span>

5 0
3 years ago
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