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Radio stations use electromagnetic waves for broadcasting. The chart shows different frequencies of waves used by radio stations

nlexa [21]

Radio Station W as the slower the frequency the longer the wave length

6 0

3 years ago

Read 2 more answers

The number of employees for a certain company has been decreasing each year by 4%. If the company currently has 650 employees an

nika2105 [10]

The answer would probably be 649.96

8 0

3 years ago

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of r from a point charge, Q, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$

$PD = 135\times 10^3\text{ V} = 135\text{ kV}$

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be x.

$270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]$

$10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10$

$\dfrac{1}{x} = 0$

$x = \infty$

The charge chould be moved to infinity

7 0

3 years ago

A woman launches a boat from one shore of a straight river and wants to land at the point directly on the opposite shore. If the

I am Lyosha [343]

Answer:

If she stands on the North side of a river flowing to the East at 5 mph,

she must head towards the SouthWest to arrive on the South side of the river directly across from her starting point and we have

x^2 + 5^2 = 10^2 where x is her speed directly across the river

x = (75)^1/2 = 8.66 mph towards the South

sin theta = 5 / 10 = 1/2

She must angle the boat at 30 deg from straight South

4 0

2 years ago

Red laser light from a He-Ne laser (λ = 632.8 nm) creates a second-order fringe at 53.2° after passing through the grating. What

Svetlanka [38]

Explanation:

It is given that,

Wavelength of red laser light, $\lambda=632.8\ nm=632.8\times 10^{-9}\ m$

The second order fringe is formed at an angle of, $\theta=53.2^{\circ}$

For diffraction grating,

$d\ sin\theta=n\lambda$

$d=\dfrac{n\lambda}{sin\theta}$ , n = 2

$d=\dfrac{2\times 632.8\times 10^{-9}}{sin(53.2)}$

$d=1.58\times 10^{-6}\ m$

The wavelength λ of light that creates a first-order fringe at 22 is given by :

$\lambda=d\ sin\theta$

$\lambda=1.58\times 10^{-6}\ sin(22)$

$\lambda=5.91\times 10^{-7}\ m$

$\lambda=591\ nm$

Hence, this is the required solution.

6 0

4 years ago