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Ad libitum [116K]
3 years ago
6

What will be the fermi temperature, if the number density is increased e a) Remains same b) Decreases by a factor of 4 c) Increa

ses by a factor of 4 d) None of the options
Physics
1 answer:
vichka [17]3 years ago
8 0

Answer:

d) None of the options

Explanation:

Fermi temperature is directly proportional to number of density.

The relationship between Fermi temperature and number density given as follows

n\alpha T^{\frac{3}{2}}

So When Fermi temperature increases then the number density will also increases.

So the option d is right because ,in the given option there is no any option which said that the number density will increase.

 

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Two tuning forks are played at the same time. One has a frequency of 490 Hz and the other is 488 Hz. How many beats per second a
Gnoma [55]
Hope this helps you.

7 0
3 years ago
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An object falls from the top of a building that is 25 m high. Air resistance is negligible.
Vlada [557]

The velocity of the object s calculated as 22.1 m/s.

<h3>What is the speed of the object?</h3>

Given that we can write that;

v^2 = u^2 + 2gh

Now u = 0 m/s because the object was dropped from a height

v^2 = 2gh

v = √2 * 9.8 * 25

v = 22.1 m/s

Learn more about velocity:brainly.com/question/18084516

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5 0
2 years ago
why does the velocity in the horizontal direction remain constant while the velocity in the vertical direction changes?
Bond [772]
The horizontal velocity<span> of a projectile is </span>constant<span> (a never </span>changing<span> in value), There is a </span>vertical<span>acceleration caused by gravity; its value is 9.8 m/s/s, down, The </span>vertical velocity<span> of a projectile </span>changes<span> by 9.8 m/s each second, The </span>horizontal<span> motion of a projectile is independent of its </span>vertical<span> motion.</span>
8 0
3 years ago
Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f
vampirchik [111]

Answer:

\large \boxed{\text{30 rev/s}}

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

 I₁ = 3.5    kg·m²s⁻¹

ω₁ = 6.0    rev·s⁻¹

 I₂ = 0.70 kg·m²s⁻¹

Calculation:

\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}

8 0
3 years ago
From an h = 53 feet observation tower on the coast, a Coast Guard officer sights a boat in difficulty. The angle of depression o
maksim [4K]

Answer:

757,93 feets

Explanation:

We can make a right triangle between the boat (A), the Coast Guard officer (B) and the base of the observation tower (C), like in the graph attached. Now, you could also made a rectangle, adding the horizontal at the height of the Coast Guard, starting in B and ending in D, the vertex opossing C.

The angle of depression, its O in the graph.

Now, as we got an rectangle, of course, the segment AD its the same length as CB, and CA, the distance from the boat to shoreline, its the same length as DB.

ADB its an right triangle, with AB, the hypothenuse, and BD and DA, the catheti (or <em>legs</em>).

Now, we know the lenght BC, the height of the tower, 53 feets, so we also know the lenght of DA. DA its the opposite cathetus to the angle O. We wish to know the length AC, equal to the lenght DB, the adjacent cathetus of the angle O.

Know, the trigonometric function that connects the adjacent cathetus with the opossite cathetus its the tangent.

tangent( O ) = \frac{opposite}{adjacent}

We can take that the angle O = 4 °, and knowing that the opossite cathetus its 53 feets, we got:

tangent( 4) = \frac{53 feets}{DB}

DB=  \frac{53 feets}{tangent( 4)}

DB=  757,93 feets

This its equal to the distance from the boat to the shoreline.

4 0
3 years ago
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