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lidiya [134]
2 years ago
8

A plane electromagnetic wave, with wavelength 5 m, travels in vacuum in the positive x direction with its electric vector E, of

amplitude 229.1 V/M, directed along y axis. What is the time-averaged rate of energy flow in watts per square meter associated with the wave
Physics
1 answer:
Dafna1 [17]2 years ago
5 0

Answer:

Explanation:

E₀ = 229.1 V/m

E = E₀ / √2 = 229.1 / 1.414 = 162 V/m

B =  E / c  ( c is velocity of em waves )

=   162 / (3 x 10⁸)  = 54 x 10⁻⁸ T

rate of energy flow =  ( E x B )  / μ₀

= 162 x 54 x 10⁻⁸ / 4π x 10⁻⁷

= 69.65 W per m².

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The valve in the exit pipe is closed . The density of water is 1000kg / m and the gravitational force on free fall of water is 1
Stels [109]

Answer:

300000

Explanation:

p=30x10x1000=30000pascal

7 0
2 years ago
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As part of a safety investigation, two 1900 kg cars traveling at 20 m/s are crashed into different barriers. Part A Find the ave
DedPeter [7]

Answer:

-29.2\times 10^{3} N

Explanation:

We are given that

Mass of cars= m=1900 kg

Initial speed of car=u=20 m/s

Final speed of car=v=0

Time=\Delta t=1.3 s

We have to find the average force exerted on the car.

Average force=\frac{change\;in\;momentum}{\Delta t}

F_{avg}=\frac{mv-mu}{1.3}

F_{avg}=\frac{1900(0)-1900(20)}{1.3}

F_{avg}=\frac{-38000}{1.3}=-29.2\times 10^{3} N

Hence, the average force exerted on the car that hits a line of water barrels=-29.2\times 10^{3} N

8 0
3 years ago
Carbon burns with oxygen to produce carbon dioxide gas. which of these shows the correct chemical reaction?
Ksju [112]
B

Chemical equations always have an arrow
4 0
3 years ago
Read 2 more answers
If the initial velocity of a ball is sent straight upward at 10.5m/s from the ground what will its final velocity be when it hit
Shalnov [3]

Answer: -10.08 m/s

Explanation:

Here we only need to analyze the vertical problem.

When the ball is in the air, the only force acting on it will be the gravitational force, this means that the acceleration of the ball, is equal to the gravitational acceleration, then:

a(t) = -9.8m/s^2

Where the negative sign is because gravity pulls the ball down.

To get the velocity equation we need to integrate over time, we get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity, here it is v0 = 10.5 m/s

Then the velocity equation is:

v(t) =  (-9.8m/s^2)*t + 10.5 m/s

To get the position equation, we need to integrate again over time, we get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t + p0

Where p0 is the initial position, we know that the ball is sent upward from the ground, so p0 = 0m

Then the position equation is:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

Now we need to find the value of t such that the position is equal to zero (this means that the ball hits the ground again).

Then we need to solve:

p(t) = 0 =  (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

If we divide both sides by t, we get:

0 =   (1/2)*(-9.8m/s^2)*t + (10.5 m/s)

Now we can solve it:

(1/2)*(9.8m/s^2)*t = 10.5 m/s

t = (10.5 m/s)/((1/2)*(9.8m/s^2)) = 2.14 s

This means that after 2.14 seconds, the ball will hit the ground again.

The velocity of the ball when it hits the ground is equal to:

v(2.14s) = (-9.8m/s^2)*2.14s + 10.5 m/s = -10.08 m/s

3 0
3 years ago
Please help on this one?
scoray [572]

option B open system

because in open system energy and mass can escape from the system or can be added to it.

8 0
3 years ago
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