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4 years ago

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A friend is telling you about a type of high-energy, short-wavelength electromagnetic wave. She is MOST LIKELY to be talking abo

ut _____.
A. X-rays
B. visible light
C. radio waves
D. infrared light

1 answer:

Masteriza [31]4 years ago

4 0

Answer: A. X-rays

Explanation:

X-rays have the shortest wavelength and the highest energy of all of the options given.

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Review. (c) Assume the dipole is a compass needle-a light bar magnet-with a magnetic moment of magnitude μ . It has moment of in

hammer [34]

The frequency of oscillation is $\frac{1}{2\pi } \sqrt{\frac{\mu B}{I} }$ .

<h3>What is a magnetic moment?</h3>

The magnetic moment is the magnetism of a magnet or other item that creates a magnetic field, as well as its orientation and strength. Electromagnets, permanent magnets, elementary particles like electrons, different compounds, and a variety of celestial objects are examples of things that have magnetic moments (such as many planets, some moons, stars, etc). The phrase "magnetic moment" typically refers to a system's magnetic dipole moment, which can be represented by an analogous magnetic dipole, which has a magnetic north and south pole that are barely separated from one another. For sufficiently small magnets or at sufficiently wide distances, the magnetic dipole component is adequate. For extended objects, additional terms may be required in addition to the dipole moment, such as the magnetic quadrupole moment.

$T = \frac{ Id^{2}\theta }{dt^{2} }$

$-\mu B\theta=\frac{ Id^{2}\theta }{dt^{2} }$

$\frac{d^{2}\theta }{dt^{2} } = -(\frac{\mu BI}{\theta} )$

By Comparing the above equation with the SHM equation

$\frac{d^2 \theta} {dt^{2} } = -\omega^{2} \theta$

$\omega^{2} =\frac{ \mu B}{I}$

Frequency = $\frac{\mu}{2\pi }$

= $\frac{\sqrt{\frac{\mu B}{I} } }{2\pi}$

= $\frac{1}{2\pi } \sqrt{\frac{\mu B}{I} }$

To learn more about a magnetic moment, visit:

brainly.com/question/17000031

#SPJ4

8 0

1 year ago

Calculate the entropy change that occurs when 1.0kg of water at 20.00 C is mixed with 2.0kg of water at 80.00 C

SOVA2 [1]

Answer:

The change in entropy ΔS = 0.0011 kJ/(kg·K)

Explanation:

The given information are;

The mass of water at 20.0°C = 1.0 kg

The mass of water at 80.0°C = 2.0 kg

The heat content per kg of each of the mass of water is given as follows;

The heat content of the mass of water at 20.0°C = h₁ = 83.92 KJ/kg

The heat content of the mass of water at 80.0°C = h₂ = 334.949 KJ/kg

Therefore, the total heat of the the two bodies = 83.92 + 2*334.949 = 753.818 kJ/kg

The heat energy of the mixture =

1 × 4200 × (T - 20) = 2 × 4200 × (80 - T)

∴ T = 60°C

The heat content, of the water at 60° = 251.154 kJ/kg

Therefore, the heat content of water in the 3 kg of the mixture = 3 × 251.154 = 753.462

The change in entropy ΔS = ΔH/T = (753.818 - 753.462)/(60 + 273.15) = 0.0011 kJ/(kg·K).

8 0

3 years ago

A bicycle is heading West. It goes 5000m in 500s what's its velocity

dybincka [34]

There are several information's of immense importance already given in the question. Based on the given information's the answer to the question can easily be determined.
Distance covered by the bicycle = 5000 meter
Time taken by the bicycle to reach the distance = 500 second.
Velocity of the bicycle = Distance / Time taken
= 5000/500 meter/second
= 50 meter/second
So the velocity of the bicycle is 50 meter per second. I hope the procedure is clear enough for you to understand. In future you can always use this procedure for solving similar problems.

8 0

3 years ago

A student produces a wave in a long spring by

Sliva [168]

Answer: 2

Explanation:

4 0

3 years ago

Show that the expression pRT has the same units as the pressure, and thus that the ideal gas law is dimensionally consistent.

Gemiola [76]

Answer and Explanation:

We have given the expression $\rho RT$

Dimension of $\rho =ML^{-3}$

Dimension of R which is gas constant $=ML^2T^{-2}\Theta ^{-1}M^{-1}$

Dimension of temperature T $\Theta ^{-1}$

And dimension of pressure $ML^{-1}T^{-2}$

Now combine dimension of $\rho RT$ $=ML^{-3}ML^2T^{-2}\Theta ^{-1}M^{-1}\Theta ^{-1}=ML^{-1}T^{-2}$

So the dimension of $\rho RT$ and dimension P is same so there unit will also be same

From ideal gas equation we know that $PV=nRT$

$P=\frac{n}{V}RT=\rho RT$

As the both P and $\rho RT$ has same dimension so they are dimensionally constant

7 0

4 years ago