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Varvara68 [4.7K]
3 years ago
6

Light waves are electromagnetic waves that travel at 3.00 Light waves are electromagnetic waves that travel 108 m/s. The eye is

most sensitive to light having a wavelength of 5.50 Light waves are electromagnetic waves that travel 10-7 m.(a) Find the frequency of this light wave.Hz(b) Find its period.s
Physics
1 answer:
svlad2 [7]3 years ago
5 0

(a) 5.45 \cdot 10^{14} Hz

The relationship between frequency and wavelength of an electromagnetic wave is given by

c=f \lambda

where

c=3.00 \cdot 10^8 m/s is the speed of light

f is the frequency

\lambda is the wavelength

In this problem, we are considering light with wavelength of

\lambda=5.50 \cdot 10^{-7} m

Substituting into the equation and re-arranging it, we can find the corresponding frequency:

f=\frac{c}{\lambda}=\frac{3.00 \cdot 10^8 m/s}{5.50 \cdot 10^{-7} m}=5.45 \cdot 10^{14} Hz

(b) 1.83\cdot 10^{-15} s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

And using f=5.45 \cdot 10^{14} Hz as we found in the previous part, we can find the period of this wave:

T=\frac{1}{5.45 \cdot 10^{14} Hz}=1.83\cdot 10^{-15} s

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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

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Answer:

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Explanation:

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This force is due to friction force when car is moving in circle with uniform speed

Now it is given that car is moving on the ice surface such that the friction force is zero now

so here we can say that centripetal force is due to component of the normal force which is due to banked road

Now we have

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so we have

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