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expeople1 [14]
3 years ago
8

What are the (time varying) amplitudes of the E and H fields if summer sunlight has an intensity of 1150 W/m2 in any Town?

Physics
1 answer:
Archy [21]3 years ago
3 0

Explanation:

Given that,

Intensity = 1150 W/m²

(a). We need to calculate the magnetic field

Using formula of intensity

I=\dfrac{E^2}{2\mu_{0}c}

E=\sqrt{2\times I\mu_{0}c}

Put the value into the formula

E=\sqrt{2\times1150\times4\pi\times10^{-7}\times3\times10^{8}}

E=931.17\ N/C

Using formula of magnetic field

B = \dfrac{E}{c}

Put the value into the formula

B=\dfrac{931.17}{3\times10^{8}}

B=0.0000031039\ T

B=3.10\times10^{-6}\ T

(b). The relative strength of the gravitational and solar electromagnetic pressure forces of the sun on the earth

We need to calculate the gravitational force

Using gravitational force

F=\dfrac{Gm_{s}M_{e}}{r^2}

Put the value into the formula

F=\dfrac{6.67\times10^{-11}\times1.98\times10^{30}\times5.97\times10^{24}}{(1.496\times10^{11})^2}

F=3.522\times10^{22}\ N

We need to calculate the radiation force

Using formula of force

F_{R}=\dfrac{I}{c}\pi\timesR_{E}^{2}

Put the value into the formula

F_{R}=\dfrac{1150}{3\times10^{8}}\times\pi\times(6.378\times10^{6})^2

F_{R}=4.8\times10^{8}\ N

The gravitational and solar electromagnetic pressure forces of the sun on the earth

\dfrac{F_{G}}{F_{R}}=\dfrac{3.522\times10^{22}}{4.8\times10^{8}}

\dfrac{F_{G}}{F_{R}}=7.3375\times10^{13}

Hence, This is the required solution.

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3 years ago
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A ball is thrown from a rooftop with an initial downward velocity of magnitude vo = 2.9 m/s. The rooftop is a distance above the
Step2247 [10]

Answer:

a) The velocity of the ball when it hits the ground is -20.5 m/s.

b) To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

Explanation:

I´ve found the complete question on the web:

<em />

<em>A ball is thrown from a rooftop with an initial downward velocity of magnitude v0=2.9 m/s. The rooftop is a distance above the ground, h= 21 m. In this problem use a coordinate system in which upwards is positive.</em>

<em>(a) Find the vertical component of the velocity with which the ball hits the ground.</em>

<em>(b) If we wanted the ball's final speed to be exactly 27, 3 m/s from what height, h (in meters), would we need to throw it with the same initial velocity?</em>

<em />

The equation of the height and velocity of the ball at any time "t" are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the ball at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the ball at a time "t".

First, let´s find the time it takes the ball to reach the ground (the time at which h = 0)

h = h0 + v0 · t + 1/2 · g · t²

0 = 21 m - 2.9 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 1.8 s  ( the other solution of the quadratic equation is rejected because it is negative).

Now, using the equation of velocity, let´s find the velocity of the ball at

t = 1.8 s:

v = v0 + g · t

v = -2.9 m/s - 9.8 m/s² · 1.8 s

v = -20.5 m/s

The velocity of the ball when it hits the ground is -20.5 m/s.

b) Now we have the final velocity and have to find the initial height. Using the equation of velocity we can obtain the time it takes the ball to acquire that velocity:

v = v0 + g · t

-27.3 m/s = -2.9 m/s - 9.8 m/s² · t

(-27.3 m/s + 2.9 m/s) / (-9.8 m/s²) = t

t = 2.5 s

The ball has to reach the ground in 2.5 s to acquire a velocity of 27.3 m/s.

Using the equation of height, we can obtain the initial height:

h = h0 + v0 · t + 1/2 · g · t²

0 = h0 -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

-h0 = -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

h0 = 38 m

To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

6 0
3 years ago
A ball is spun around in circular motion such that it completes 50 rotations in 25 s.
Leokris [45]

Answer:

(A) The period of its rotation is 0.5 s (2) The frequency of its rotation is 2 Hz.

Explanation:

Given that,

a ball is spun around in circular motion such that it completes 50 rotations in 25 s.

(1). Let T be the period of its rotation. It can be calculated as follows :

T=\dfrac{25}{50}\\\\T=0.5\ s

(2). Let f be the frequency of its rotation. It can be defined as the number of rotations per unit time. So,

f=\dfrac{50}{25}\\\\f=2\ Hz

Hence, this is the required solution.

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Water flowing from a waterfall before it hits the pond below?
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Answer:

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Explanation:

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Explanation:

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6 0
3 years ago
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