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daser333 [38]
3 years ago
12

The average Intensity of the electromagnetic field due to sunlight at the surface of the earth is 1400 W/m^2. Calculate the maxi

mum values of the electric field and magnetic field.
Physics
1 answer:
Ann [662]3 years ago
4 0

Answer:

1027 N/C

3.42 x 10⁻⁶ T

Explanation:

I = Intensity of electromagnetic field = 1400 W/m²

E₀ = Maximum value of electric field

Intensity of electromagnetic field is given as

I = (0.5) ε₀ E₀² c

1400 =  (0.5) (8.85 x 10⁻¹²) (3 x 10⁸) E₀²

E₀ = 1027 N/C

B₀ = maximum value of magnetic field

using the equation

E₀ = B₀ c

1027 = B₀ (3 x 10⁸)

B₀ = 3.42 x 10⁻⁶ T

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. A 40.0-kg child standing on a frozen pond throws a 0.500-kg stone to the east with a speed of 5.00 m/s. Neglecting friction be
tamaranim1 [39]

Answer:

Explanation:

A 40kg child throw stone of 0.5kg

At a direction of 5m/s

Recoil can be calculated using recoil of a gun formula

m_1•v_1 + m_2•v_2

m_1•v_1 = -m_2•v_2

The negative sign show that the momentum of the boy is directed oppositely to that of the stone

m_1 Is mass of boy

v_1 is the recoil velocity of the boy

m_2 is mass of stone

v_2 is the velocity of stone

Then,

m_1•v_1 = -m_2•v_2

40•v_1 = -0.5 × 5

40•v_1 = -2.5

v_1 = -2.5 / 40

v_1 = -0.0625 m/s

The recoil velocity of the boy is 0.0625 m/s

6 0
3 years ago
A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 8.00 at a distance of 4.00 fro
slavikrds [6]

Answer:

A. We have that radius r = 4.00m intensity I = 8.00 W/m^

total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W

b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2

c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J

5 0
3 years ago
A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend
Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}.

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ



According to the question, we have -

Entering Velocity (v) = 20 i  m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j  m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $\frac{2\pi r}{4} = $\frac{\pi R}{2}

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $\frac{\pi R}{2v} = \frac{\pi R}{40}

Hence, the time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

#SPJ4



 



6 0
1 year ago
What is meaning of phase​
algol [13]

Answer:

phares

phares \\  \\ means \: a \: noun \:  \\  \\ i \: thnk \: so \:  \:

5 0
2 years ago
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