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2 years ago

14

What is the term used to label the energy levels of electrons?

1 answer:

mel-nik [20]2 years ago

4 0

Electron volts...........

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A private aviation helicopter's main rotor blades rotate at approximately

Arisa [49]

Answer: 7.5 rev/s

Explanation:

We are given the angular velocity $\omega$ a helicopter's main rotor blades:

$\omega=450 rpm=450 \frac{rev}{min}$

However, we are asked to express this $\omega$ in the International Systrm (SI) units. In this sense, the SI unit for time is second ( $s$ ):

$\omega=450 \frac{rev}{min} \frac{1 min}{60 s}$

$\omega=7.5 \frac{rev}{s}$

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2 years ago

Your skeleton. It is there for support, protection, and movement. But you could not move your bones without the help of your ___

lana66690 [7]

I believe the answer is D.) the muscular system

7 0

2 years ago

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What's the gravity?

poizon [28]

Answer:

gravity is the force by which a planet or another body draws objects towards its Centre. the force of gravity keeps all of the planets in an orbit around the sun.

Explanation:

what is gravity made of?

gravity are made of proposed gravity parts that are made quantum particles which called gravitational anywhere there is gravitation there is gravity like on Earth in the Solar System most importantly in the minisclue infant and intervention universe .

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2 years ago

Which illustration represents low accuracy but high precision?

Misha Larkins [42]

Answer:

D i think

Explanation:

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3 years ago

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We want to design a cylindrical vacuum capacitor, with a given radius a for the outer cylindrical shell, that will be able to st

anastassius [24]

Solution :

a). Using Gauss's law :

$E=\frac{Q}{4 \pi \epsilon_0r^2}$ , $$b$ .........(1)

Let $E=E_0,\ r=b$ in equation (1)

Therefore, $Q=4 \pi \epsilon_0b^2E_0$ .............(2)

$V_b-V_a = \int^a_b \vec E. d\vec l$

$=\int^a_b E \ dx$

$=\frac{Q}{4 \pi \epsilon_0} \int^a_b \frac{1}{x^2} \ dx$

$=\frac{Q(a-b)}{4 \pi \epsilon_0 a b}$ ....................(3)

Therefore, $U=\frac{1}{2}Q \Delta V$

$=\frac{1}{2}(4 \pi \epsilon_0 b^2 E_0)\left(\frac{Q(a-b)}{4\pi \epsilon_0 a b}\right)$

$=\frac{4 \pi \epsilon_0}{2a} \ E^2_0 b^3(a-b)$ .............(4)

Now differentiating the equation (4) w.r.t. 'b', we get

$b=\frac{3}{4}a$

Thus the radius for the inner cylinder conductor is $b=\frac{3}{4}a$

b). For the energy storage, substitute the radius in (4), we get

$U = 4 \pi\epsilon_0 \frac{27a^3E^2_0}{512}$

This is the amount of energy stored in the conductor.

8 0

2 years ago