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2 years ago

8

One of the frequencies used to transmit and receive cellular telephone signals in the United States is 985 MHz. What is the wave

length in meters of these radio waves

1 answer:

tia_tia [17]2 years ago

5 0

Answer:

approximately 304358

Explanation:

wavelength = speed / frequency

speed of an electromagnetic wave is 299,792,458 m/s in a vacuum

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A certain dielectric with a dielectric constant = 24 can withstand an electric field of 4 107 V/m. Suppose we want to use this d

mr_godi [17]

Answer:

Explanation:

Electric field E = 4 x 10⁷ V / m

Dielectric constant k = 24

capacitance of capacitor

C = kε₀ A / d

d = plate separation

A = plate area

C = .89 x 10⁻⁶

V / d = electric field

for minimum d , electric field will be maximum

V / d = 4 x 10⁷

1930 / d = 4 x 10⁷

d = 1930 / 4 x 10⁷

d = 482.5 x 10⁻⁷ m

= 48.25 x 10⁻⁶ m

C = kε₀ A / d

.89 x 10⁻⁶ = 24 ε₀ A / d

A = .89 x 10⁻⁶ X d / 24 ε₀

A = .89 x 10⁻⁶ X 48.25 x 10⁻⁶ / 24 x 8.85 x 10⁻¹²

= 42.9 / 212.4

= .2019 m²

8 0

3 years ago

Read 2 more answers

Excellent human jumpers can leap straight up to a height of

Firlakuza [10]

For a human jumper to reach a height of 110 cm, the person will need to leave the ground at a speed of 4.65 m/s.

We can calculate the initial speed to reach 110 cm of height with the following equation:

$v_{f}^{2} = v_{i}^{2} - 2gh$

Where:

$v_{f}$ : is the final speed = 0 (at the maximum height of 110 cm)

$v_{i}$ : is the initial speed =?

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 110 cm = 1.10 m

Hence, the <u>initial velocity</u> is:

$v_{i} = \sqrt{v_{f}^{2} + 2gh} = \sqrt{2*9.81 m/s^{2}*1.10 m} = 4.65 m/s$

Therefore, the initial speed that the person must have to reach 110 cm is 4.65 m/s.

You can see another example here: brainly.com/question/13359681?referrer=searchResults

I hope it helps you!

4 0

2 years ago

For what absolute value of the phase angle does a source deliver 71 % of the maximum possible power to an RLC circuit

kherson [118]

Answer: the absolute value of the phase angle is 28°

Explanation:

taking a look at expression for the instantaneous electric power in an AC circuit;

P = VI -------let this be equation 1

p is power, v is voltage and I is current;

for maximum power

P_max = V_rms × I_rms --------let this be equ 2

where P_max is the maximum power, V_rms is the rms value of voltage and I_rms is the rms value of current.

Also for average electric power in an AC circuit

P_avg = V_rms × I_rms × cos²∅ -------let this be equ 3

where P_avg is the average power and cos∅ is the power factor

now from equation 2; P_max = V_rms × I_rms

so p_max replaces V_rms × I_rms in equation 3

we now have

P_avg = P_max × cos²∅

so we substitute

expression for the given value of the average power is

P_avg = P_max × 75%

p_avg = P_max.78/100

for the expression of the average electricity in an AC circuit

P_max.78/100 = P_max × cos²∅

78/100 = cos²∅

to get the absolute value of phase angle

∅ = cos⁻¹ ( √(78/100))

∅ = cos⁻¹ ( 0.8832)

∅ = 27.969 ≈ 28°

Therefore the absolute value of the phase angle is 28°

8 0

2 years ago

En el mar, la luz visible alcanza una profundidad de aproximadamente 200 metros (zona fotica). ¿De dónde obtienen energías las p

Marizza181 [45]

Answer:

Explanation:

En la zona apótica (profundidad inferior a 200 m); todo lo que queda de la luz solar es una luz tenue, opaca, azul-verde, demasiado impotente para siquiera considerar permitir que ocurra la fotosíntesis. Sin embargo, hay comida para tener; basura, trozos de plantas podridas y derroche de criaturas caen desde arriba para cuidar a los seres vivos en la zona apótica.

Las formas de vida a una profundidad inferior a 200 m dependen de los productos químicos que salen de los respiraderos; el procedimiento que utilizan para hacer los alimentos se llama quimiosíntesis en lugar de fotosíntesis.

3 0

2 years ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago