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1 year ago

Place the following types of electromagnetic radiation in order from LOWEST energy

Physics

1 answer:

Alex1 year ago

3 0

Placing the electromagnetic radiation in order from the lowest energy to the highest energy : ( 2 ) C,A,B

<h3>Electromagnetic spectrum </h3>

In the electromagnetic spectrum the electromagnetic radiation with the shorter wavelength possess a higher energy while the electromagnetic radiation with a longer wavelength possess the lower energy.

The electromagnetic radiation as listed in the question with the longest wavelength is the radio waves therefore it possess the lowest energy while the radiation with the shortest wavelength is the gamma rays therefore it possess the highest energy.

Hence we can conclude that Placing the electromagnetic radiation in order from the lowest energy to the highest energy : ( 2 ) C,A,B

Learn more about electromagnetic spectrum : brainly.com/question/25847009

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Which of the three recording stations was closest to the epicenter of the earthquake, based on the seismograms shown below?

Nastasia [14]

Answer:C

Explanation:took quiz

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3 years ago

Read 2 more answers

Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh

frosja888 [35]

Answer:

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

Explanation:

Physically speaking, the resulting velocity of the helicopter ( $\vec v_{H}$ ), measured in meters per second, is equal to the absolute velocity of the wind ( $\vec v_{W}$ ), measured in meters per second, plus the velocity of the helicopter relative to wind ( $\vec v_{H/W}$ ), also call velocity at still air, measured in meters per second. That is:

$\vec v_{H} = \vec v_{W}+\vec v_{H/W}$ (1)

In addition, vectors in rectangular form are defined by the following expression:

$\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha)$ (2)

Where:

$\|\vec v\|$ - Magnitude, measured in meters per second.

$\alpha$ - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

$\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W})$ (3)

$\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)$

If we know that $\|\vec v_{W}\| = 25\,\frac{m}{s}$ , $\|\vec v_{H/W}\| = 125\,\frac{m}{s}$ , $\alpha_{W} = 240^{\circ}$ and $\alpha_{H/W} = 325^{\circ}$ , then the resulting velocity of the helicopter is:

$\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)$ $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

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3 years ago

Measuring the volume of a ball that is 24cm across how can you set up an equation

BigorU [14]

The volume of every sphere is

   Volume = (4/3) (pi) (radius)³

When you say "across", I think you mean the diameter of the ball.
The radius is half of the diameter = 12 inches.

   Volume = (4/3) (pi) (12 inches)³

   = (4/3) (pi) (1,728 cubic inches)

   =    7,238.2 cubic inches . (rounded)

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3 years ago

Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric

Goshia [24]

Complete Question:

Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric force on sphere B due to sphere A is F . F. The magnitude of the electric force on sphere A due to sphere B must be:

A. 2F

B. F/4

C. F/2

D. F

E. 4F

Answer:

Explanation:

If both spheres can be treated as point charges, they must obey the Coulomb's law, that can be written as follows (in magnitude):

$F =\frac{kQ*2Q}{r^{2} }$

As it can be seen, this force is proportional to the product of the charges, so it must be the same for both charges.

As this force obeys also the Newton's 3rd Law, we conclude that the magnitude of the electric force on sphere A due to sphere B, must be equal to the the magnitude of the force on the sphere B due to the sphere A, i.e., just F.

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3 years ago

When two light waves arrive at the same place at the same time they create a?

Montano1993 [528]

Hey there,
<span>They interfere essentially like any other form of wave.
</span>
Hope this helps :))

~Top

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3 years ago