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Setler79 [48]
3 years ago
10

the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic f

ield for that wave at P
Physics
1 answer:
vesna_86 [32]3 years ago
7 0

Answer:

6.63\times 10^8\ N/C

Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

We need to find the magnitude of the electric field. Let it is E. So,

\dfrac{E}{B}=c\\\\E=Bc

Put all the values,

E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C

So, the magnitude of the electric field is equal to 6.63\times 10^8\ N/C.

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A sprinter must average 24.0 mi/h to win a 100-m dash in 9.30 s. What is his wavelength at this speed if his mass is 84.5 kg?
crimeas [40]

Answer:

Wavelength λ = 7.31 × 10^-37 m

Explanation:

From De Broglie's equation;

λ = h/mv

Where;

λ = wavelength in meters

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m = mass in kg

v = velocity in m/s

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v = 24 mi/h

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v = 10.73m/s

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Substituting the values into the equation;

λ = (6.626×10^-34 m^2 kg/s)/(84.5kg × 10.73m/s)

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3 years ago
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eimsori [14]
Because AC emf is a sine wave
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A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
Y_Kistochka [10]

Answer:

See Explanation

Explanation:

a) We know that;

v = λf

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f = frequency of the wave

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So;

T = 2 * 2.10 s = 4.2 s

Hence f = 1/4.2 s

f = 0.24 Hz

The wavelength =  6.5 m

Hence;

v = 6.5 m * 0.24 Hz

v = 1.56 m/s

b)The amplitude of the wave is;

A =  0.600 m/2 = 0.300 m

c) Since the wave speed does not depend on the amplitude of the wave then the answer in (a) above remains the same

Where d = 0.30 m

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3 years ago
A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik
shutvik [7]

Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

Launching angle=40^{\circ}

x =12\pm 0.27

y=1.05

equation of trajectory of ball is given by

y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }

for x=12.27

1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }

u=10.69

for x=11.73

1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }

u=11.436 m/s

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The answer is d. functionalist perspective 
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