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irinina [24]
3 years ago
9

Cell phone conversations are transmitted by high-frequency radio waves. Suppose the signal has wavelength 36.5 cm while travelin

g through air. What is the frequency as the signal travels through 3-mm-thick window glass into your room?
Physics
1 answer:
tia_tia [17]3 years ago
7 0

Answer:

f=8.219*10^{8}Hz

Explanation:

We are going to use the formula  v=fλ

Where v= velocity of radio waves

f= frequency

λ= wavelength of wave

  • radio waves are electromagnetic waves and as such they have the speed of light which is 3*10^{8}m/s.
  • also when a wave travels from one medium to another, the wavelength changes while the frequency remains the same.
  • calculating for the frequency of the wave in air also gives us the frequency in the window glass.

f=\frac{v}{λ}

v=3*10^{8}m/s.

λ=36.5 cm = 36.5/100= 0.365m

f=\frac{3*10^{8}m/s.}{0.365m}

f=8.219*10^{8}Hz

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Un movil viaja a 40km/h y comienza a reducir su velocidad a partir del instante t=0. Al cabo de 6 segundo se detiene completamen
aleksklad [387]

Answer:

1,85 m / s²

Explanation:

De la pregunta anterior, se obtuvieron los siguientes datos:

Velocidad inicial (u) = 40 km / h

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:

1 km / h = 0,2778 m / s

Por lo tanto,

40 km / h = 40 km / h × 0,2778 m / s / 1 km / h

40 km / h = 11,11 m / s

Por tanto, 40 km / h equivalen a 11,11 m / s.

Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:

Velocidad inicial (u) = 11,11 m / s

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

a = (v - u) / (t₂ - t₁)

a = (0 - 11,11) / (6 - 0)

a = - 11,11 / 6

a = –1,85 m / s²

Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²

6 0
3 years ago
. During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from th
mixer [17]

Answer:

106.7 N

Explanation:

We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:

F \Delta t = m (v-u)

where

F is the average force

\Delta t is the duration of the collision

m is the mass of the ball

v is the final velocity

u is the initial velocity

In this problem:

m = 0.200 kg

u = 20.0 m/s

v = -12.0 m/s

\Delta t = 60.0 ms = 0.06 s

Solving for F,

F=\frac{m(v-u)}{\Delta t}=\frac{(0.200 kg) (-12.0 m/s-20.0 m/s)}{0.06 s}=-106.7 N

And since we are interested in the magnitude only,

F = 106.7 N

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