What is one characteristic of the organization of the Periodic Table?

Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth

Semmy [17]

Answer:

$\lambda=6.83\times 10^{-5}\ m$

Explanation:

Given that,

An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 4.39 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

We know that,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m$

So, the wavelength of the infrared radiation is $6.83\times 10^{-5}\ m$ .

5 0

3 years ago

A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif

Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c) w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

μ = N*I*A

Where,

N: The number of turns

I : Current in coil

A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

vec ( τi ) = vec ( μi ) x vec ( B )

= 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

= ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

$\left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right] = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\$

- The initial torque ( τi ) is written as follows:

vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

Ui = - vec ( μi ) . vec ( B )

Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

Ui = - [(-0.000605556 + 0.00011)]

Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

Uf = - vec ( μf ) . vec ( B )

Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

ΔU = Uf - Ui

ΔU = -0.000252315 - 0.000495556

ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

Ui + Eki = Uf + Ekf

Where,

Eki : The initial kinetic energy ( initially at rest ) = 0

Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.

-ΔU = Ekf

0.5*T*w^2 = -ΔU

w^2 = -ΔU*2 / T

Where,

w: The angular speed at second position

w = √(0.000747871*2 / 6.50×10^−7)

w = 47.97 rad / s

6 0

3 years ago

Can you explain that gravity pulls us to the Earth & can you calculate weight from masses on both on Earth and other planets

schepotkina [342]

I don't actually understand what your question is, but I'll dance around the subject
for a while, and hope that you get something out of it.

-- The effect of gravity is: There's a pair of forces, in both directions, between
every two masses.

-- The strength of the force depends on the product of the masses, so it doesn't matter whether there's a big one and a small one, or whether they're nearly equal.
It's the product that counts. Bigger product ==> stronger force, in direct proportion.

-- The strength of the forces also depends on the distance between the objects' centers. More distance => weaker force. Actually, (more distance)² ==> weaker force.

-- The forces are equal in both directions. Your weight on Earth is exactly equal to
the Earth's weight on you. You can prove that. Turn your bathroom scale face down
and stand on it. Now it's measuring the force that attracts the Earth toward you.
If you put a little mirror down under the numbers, you'll see that it's the same as
the force that attracts you toward the Earth when the scale is right-side-up.

-- When you (or a ball) are up on the roof and step off, the force of gravity that pulls
you (or the ball) toward the Earth causes you (or the ball) to accelerate (fall) toward the Earth.
Also, the force that attracts the Earth toward you (or the ball) causes the Earth to accelerate (fall) toward you (or the ball).
The forces are equal. But since the Earth has more mass than you have, you accelerate toward the Earth faster than the Earth accelerates toward you.

-- This works exactly the same for every pair of masses in the universe. Gravity
is everywhere. You can't turn it off, and you can't shield anything from it.

-- Sometimes you'll hear about some mysterious way to "defy gravity". It's not possible to 'defy' gravity, but since we know that it's there, we can work with it.
If we want to move something in the opposite direction from where gravity is pulling it, all we need to do is provide a force in that direction that's stronger than the force of gravity.
I know that sounds complicated, so here are a few examples of how we do it:
-- use arm-muscle force to pick a book UP off the table
-- use leg-muscle force to move your whole body UP the stairs
-- use buoyant force to LIFT a helium balloon or a hot-air balloon
-- use the force of air resistance to LIFT an airplane.

-- The weight of 1 kilogram of mass on or near the Earth is 9.8 newtons. (That's
about 2.205 pounds). The same kilogram of mass has different weights on other planets. Wherever it is, we only know one of the masses ... the kilogram. In order
to figure out what it weighs there, we need to know the mass of the planet, and
the distance between the kilogram and the center of the planet.

I hope I told you something that you were actually looking for.

7 0

3 years ago

Water that flows from behind a large dam can cause machines to produce electricity. What change takes place?

lyudmila [28]

The mechanical energy in the falling water is used to spin the generator, and gets transformed into electrical energy. That's the first choice on the list.

8 0

3 years ago

The three basic economic questions societies face are (1) what goods and services to produce, (2) how to produce those goods and

lord [1]

Answer:

For whom are goods and services to be produced? In other words, who gets what?

What should we produce?

For whom should we produce it?

Explanation:

8 0

3 years ago