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fomenos
3 years ago
10

The magnetic field in an electromagnetic wave has a peak value given by b = 4.8 μt . part a part complete for this wave, find th

e peak electric field strength. express your answer using two significant figures. e = 1.4 kv/m submitprevious answers correct part b find the peak intensity. express your answer using two significant figures.
Physics
1 answer:
liubo4ka [24]3 years ago
6 0
The relationship between intensity of the magnetic field B and intensity of the electric field E for an electromagnetic wave is given by
E=cB
where c=3 \cdot 10^8 m/s is the speed of light.

This relationship holds also for the peak intensity of the two fields. Since the peak intensity of the magnetic field is
B=4.8 \mu T= 4.8 \cdot 10^{-6}T
we can use the previous relationship to find the peak value of the electric field:
E=(3 \cdot 10^8 m/s)(4.8 \cdot 10^{-6} T)=1440 V/m=1.4 kV/m
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4 years ago
What is Otter's average velocity over his entire trip when it takes him 2 minutes to walk 100 meters north and another 1 minute
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0.50m/s

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4 years ago
A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
bogdanovich [222]

Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

5 0
4 years ago
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