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Nady [450]
3 years ago
7

Which of the following is/are true? Electromagnetic waves are created by accelerating charges. Electromagnetic waves are transve

rse waves. Electromagnetic waves consist of time-varying electric and magnetic fields that are mutually perpendicular. Electromagnetic waves move at the speed of light in a vacuum. All of the above are true.
Physics
1 answer:
Vikki [24]3 years ago
4 0

Answer:

All of the above are true.

Explanation:

(a). true  

whenever  charge particle  move back and froth from its mean position then it will produce oscillating electric and magnetic fields, . so an em wave can be obtain by accelerating charge

(b). true

the electric field and the magnetic field have vibrations in the perpendicular direction along  the motion of the wave  so electromagnetic wave is a transverse wave. therefore, the EM wave is a Transverse wave

(c) true .

The Electromagnetic wave consists of the two mutually perpendicular electric and magnetic fields  and also both fields are  perpendicular to the direction of propagation of the wave.

(d) true .

An electromagnetic wave  carry  energy through  vacuum with a speed   of 3 \times 10^8 m/s  

   so , all of the above are true.

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a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
3 years ago
The lons entering the mass spectrometer have the same charges. After being accelerated through a potential difference of 8.20 kV
Ratling [72]

The calculated magnitude is  6.73 x 10³ V/m.

AMU is described as being one-twelfth the mass of a carbon-12 atom (12C). C makes up more than 98% of the carbon that can be found in nature, making it the most prevalent isotope. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.

Potential difference = 8.20 kV= 8.20 x 10³ V

radius= 19.4/100=0.194 m

total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194

                                                                               = 1.218 m

therefore Magnitude= 8.20 x 10³ / 1.218

                                  =6.73 x 10³ V/m

Learn more about Magnitude here-

brainly.com/question/15681399

#SPJ9

4 0
1 year ago
What are the thee major wind systems
astra-53 [7]
Trade winds, prevailing westerlies, polar easterlies
8 0
3 years ago
Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

6 0
2 years ago
Why is a minimum of three seismic stations needed to find the epicenter of an earthquake?
vazorg [7]

Each station can detect how far away the epicenter was. So each station basically has a circle made of possible epicenters. When you have three, you narrow it down to one, final point.

5 0
3 years ago
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