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iris [78.8K]
3 years ago
9

The period of a carrier wave is T=0.01 seconds. Determine the frequency and wavelength of the carrier wave. a. f=10 Hz, λ=3E8 me

ters b. f=100 Hz, λ=3E7 meters c. f=100 Hz, λ=3E6 meters d. f=10 Hz, λ=3E7 meters
Physics
1 answer:
makvit [3.9K]3 years ago
6 0

Explanation:

It is given that,

The period of the carrier wave, T = 0.01 s

Let f and \lambda are frequency and the wavelength of the wave respectively. The relationship between the time period and the frequency is given by :

f=\dfrac{1}{T}

f=\dfrac{1}{0.01}

f = 100 Hz

The wavelength of a wave is given by :

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8}{100}

\lambda=3\times 10^6\ m

So, the  frequency and wavelength of the carrier wave are 100 Hz and 3\times 10^6\ m respectively. Hence, the correct option is (c).

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Answer:

4x

Explanation:

Use v^{2} = u^{2} +2as to do the question.

For first instance,

0 = v^{2} +2ax -------------------( 1 )

for second instance,

0 = (2v)^{2} +2as-----------------( 2 )

So by (1) and (2),

s = 4x

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3 years ago
A 5.0 cm object is 12.0 cm from a concave mirror that has a focal length of 24.0 cm. The distance between the image and the mirr
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Answer:

The answer is 24cm

Explanation:

This problem bothers on the curved mirrors, a concave type

Given data

Object height h= 5cm

Object distance = 12cm

Focal length f=24cm

Let the image distance be v=?

Applying the formula we have

1/v +1/u= 1/f

Substituting our given data

1/v+1/12=1/24

1/v=1/24-1/12

1/v=1-2/24

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This implies that the image is on the same side as the object and it is real

7 0
3 years ago
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The critical angle for a liquid in air is 520, What is the liquid's index of refraction? 0.62 0.79 1.27 1.50
sammy [17]

Answer:

Liquid's index of refraction, n₁ = 1.27

Explanation:

It is given that,

The critical angle for a liquid in air is, \theta_c=52^o

We have to find the refractive index of the liquid. Critical angle of a liquid is defined as the angle of incidence in denser medium for which the angle of refraction is 90°.

Using Snell's law as :

n_1sin\theta_c=n_2sin\theta_2

Here, \theta_2=90

sin\theta_c=\dfrac{n_2}{n_1}

Where

n₂ = Refractive index of air = 1

n₁ = refractive index of liquid

So,

n_1=\dfrac{n_2}{sin\theta_c}

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n₁ = 1.269

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8 0
3 years ago
A boy pulls a sled of mass 5.0 kg with a rope that makes a 60.0° angle with respect to the horizontal surface of a frozen pond.
Oduvanchick [21]

Answer:

μk = 0.124

Explanation:

Known data

m=5.0 kg : mass of the sled

T= 10 N   : force with which the boy pulls the rope

θ =60.0°  :angle of the rope with respect to the horizontal direction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law to the sled :

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

Forces acting on the sled

W: Weight of the sled : In vertical and downward direction

N : Normal force : In vertical and upwards direction

f : Friction force: parallel to the movement of the sled and in the opposite direction to the movement

T:Rope tension : forming angle 60.0° of  of the rope with respect to the horizontal direction

Calculated of the W  of the sled

W= m*g

W=  5.0 kg* 9.8 m/s² = 49 N

x-y  components  of the tension of the rope  T

Tx= 10*cos60°= 5 N

Ty=  10*sin60° = 8.66 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N+Ty -W = 0

N = 49 N  -  8.66 N

N = 40.34 N

Calculated of the f

f = μk* N

f = μk* 40.34 Equation (1)

We apply the formula (1) to calculated f

∑Fx = m*ax  the sled moves with constant velocity, then ax=0

∑Fx = 0

Tx-f = 0

5 - f = 0

f =  5N

We replace f in the equation (1)

5 = μk* 40.34

μk = 5 / 40.34

μk = 0.124

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