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Varvara68 [4.7K]

4 years ago

5

The Sun delivers an average power of 1150 W/m2 to the top of the Earth’s atmosphere. The permeability of free space is 4π × 10−7

T · N/A and the speed of light is 2.99792 × 108 m/s. Find the magnitude of Em for the electromagnetic waves at the top of the atmosphere. Answer in units of N/C.

1 answer:

My name is Ann [436]4 years ago

4 0

Answer:

E=930.84 N/C

Explanation:

Given that

I = 1150 W/m²

μ = 4Π x 10⁻⁷

C = 2.999 x 10⁸ m/s

E= C B

C=speed of light

B=Magnetic filed ,E=Electric filed

Power P = I A

A=Area=4πr² ,I=Intensity

$I=\dfrac{CB^2}{2\mu_0}$

$I=\dfrac{CE^2}{2\mu_0 C^2}$

$E=\sqrt{{2I\mu_0 C}}$

$E=\sqrt{{2\times 1150\times 4\pi \times 10^{-7}(2.99792\times 10^8)}}$

E=930.84 N/C

Therefore answer is 930.84 N/C

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The battery capacity of a lithium ion battery in a digital music player is 750 mA-h. The manufacturer claims that the player can

Oksi-84 [34.3K]

Answer:

The number of electrons is $6.3\times10^{21}\ electrons$

(D) is correct option.

Explanation:

Given that,

Battery capacity = 750 mA-h

Time t= 8 hours

Time t'=3 hours

We need to calculate the battery capacity

$Battery\ capacity=750\times10^{-3}\times3600$

$Battery\ capacity=2700\ A-s$

We need to calculate the number of electrons in 1 C Li

Using formula for number of electron

$n=\dfrac{1}{e}$

$n=\dfrac{1}{1.6\times10^{-19}}$

$n=6.25\times10^{18}\ electrons$

We need to calculate the number of electron in 2700 C

$2700\ C=2700\times6.25\times10^{18}=1.68\times10^{22}\ electrons$

The total number of electrons battery can deliver in 8 hours

$n=1.68\times10^{22}\ electrons$

We need to calculate the number of electron in 3 hours

Using formula of number of electrons

$n=\dfrac{n\times t'}{t}$

Put the value into the formula

$n=\dfrac{1.68\times10^{22}\times3}{8}$

$n=6.3\times10^{21}\ electrons$

Hence, The number of electrons is $6.3\times10^{21}\ electrons$

8 0

4 years ago

Read 2 more answers

Harrizon [31]

Hey can you please help me?

6 0

3 years ago

A pitcher throws a 0.144-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just

Solnce55 [7]

(a) 12.8 kg m/s

The impulse delivered by the bat on the baseball is equal to the change in momentum of the baseball:

$I=\Delta p = m(v-u)$

where we have

m = 0.144 kg is the mass of the ball

v = -47 m/s is the final velocity of the ball

u = 42 m/s is the initial velocity of the ball

Substituting into the equation, we find

$I=(0.144 kg)(-47 m/s-(42 m/s))=-12.8 kg m/s$

And since we are interested in the magnitude only,

$I=12.8 kg m/s$

(b) 2.78 kN

The impulse exerted on the ball is also equal to the product between the average force and the contact time:

$I=F\Delta t$

where

F is the average force exerted on the ball

$\Delta t=0.0046 s$ is the contact time

Solving the formula for F, we find

$F=\frac{I}{\Delta t}=\frac{12.8 kg m/s}{0.0046 s}=2783 N = 2.78 kN$

(c) The force exerted on the ball is much larger (1988 times more) than the weigth of the ball

The weight of the ball is given by

$W=mg$

where

m = 0.144 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration due to gravity

Solving the equation for W, we find

$W=(0.144 kg)(9.8 m/s^2)=1.4 N$

So as we see, the force exerted on the ball (2783 N) is almost 2000 times larger than the weight of the ball (1.4 N):

$\frac{F}{W}=\frac{2783 N}{1.4 N}=1988$

8 0

3 years ago

Welcome to this IE. You may navigate to any page you've seen already using the IE Outline tab on the right. A particle beam is m

Genrish500 [490]

Answer:

the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

Explanation:

Given the data in the question;

Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J

Mass of proton = 1.673 × 10⁻²⁷ kg

Charge of proton = 1.602 × 10⁻¹⁹ C

distance d = 2 m

we know that

Kinetic Energy = Charge of proton × Potential difference ΔV

so

Potential difference ΔV = Kinetic Energy / Charge of proton

we substitute

Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )

Potential difference ΔV = 20287.14 V

Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;

E = Potential difference ΔV / distance d

we substitute

E = 20287.14 V / 2 m

E = 10143.57 V/m or 1.01 × 10⁴ V/m

Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

3 0

3 years ago

Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B

andreyandreev [35.5K]

Explanation:

Let $\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}$ and $\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}$

The sum of the two vectors is

$\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}$

$= 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}$

The difference between the two vectors can be written as

$\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}$

$= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}$

8 0

3 years ago