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wariber [46]
4 years ago
5

Find the intensity of the electromagnetic wave described in each case. (a) an electromagnetic wave with a wavelength of 630 nm a

nd a peak electric field magnitude of 7.3 v/m. w/m2 (b) an electromagnetic wave with an angular frequency of 7.7 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 t.
Physics
1 answer:
qaws [65]4 years ago
3 0

Find the intensity of the electromagnetic wave described in each case.

(a) an electromagnetic wave with a wavelength of 645 nm and a peak electric field magnitude of 8.5 V/m.
  
(b) an electromagnetic wave with an angular frequency of 6.3 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 T.

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James and Juan were at the highest point in the football stadium, dropping water balloons on people below. Many of the balloons
Fofino [41]
When you add more water to the balloon, it makes it heavier. Therefore it would weigh the balloon down ( increasing mass) and increasing the energy to plummet down. So the answer is B.
4 0
4 years ago
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A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the e
andrew-mc [135]
Average velocity = (x( 2.08 ) - x ( 0 )) / ( 2.08 s - 0 s )
x ( 2.08 ) = 1.42 * 2.08² - 0.05 * 2.08³ =
= 1.42 * 4.3264 - 0.443456 = 6.143484 - 0.443456 ≈ 5.7 m
v = ( 5.7 m - 0 m) / (2.08 s - 0 s ) = 5.7 / 2.08 m/s = 27.4 m/s
3 0
3 years ago
a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

4 0
4 years ago
Renewable energy sources are ____________________?
Nina [5.8K]

Answer:

Eco-friendly

Explanation:

6 0
3 years ago
En un rio una Onda viaja con una velocidad de propagación de 50 m/s con una longitud de Onda de 40 metros. Hallar la frecuencia
sattari [20]

Answer:

Frequencia = 1.25 Hz

Explanation:

<u>Dados los siguientes datos;</u>

  • Velocidad = 50 m/s
  • Longitud de onda = 40 metros

Para encontrar la frecuencia de la onda;

Matemáticamente, la velocidad de una onda viene dada por la fórmula;

Velocidad = Longitud \; de \; onda * Frequencia

Haciendo de la frecuencia el tema de la fórmula, tenemos;

Frequencia = \frac {Velocidad}{Longitud \; de \; onda}

Sustituyendo en la fórmula, tenemos;

Frequencia = \frac {50}{40}

<em>Frequencia = 1.25 Hz</em>

7 0
3 years ago
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