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4 years ago

In electromagnetic waves, frequency is inversely proportional to what?

1 answer:

3 0

In ANY kind of wave, frequency is inversely proportional
to wavelength.

That means their product is always the same number ...
the wave speed.

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What is the density of the material in g/cm³

Juli2301 [7.4K]

Thus 1 cm–3 = 1/cm3 and the units for our densities could be written as gcm3, g/cm3, or g cm–3. In each case the units are read as grams per cubic centimeter, the per indicating division.) We often abbreviate "cm3" as "cc", and 1 cm3 = 1 mL exactly by definition.

...

1.9: Density.

5 0

3 years ago

Which of the following quantities are vectors?

Arte-miy333 [17]

Answer:

electric fields

Explanation:

3 0

3 years ago

Nora's hair dryer is 64% efficient. If 1000 J of energy is supplied to the hair dryer, how much energy will it transfer usefully

lakkis [162]

Answer:

The hair dryer will transfer 640J of energy usefully to Nora’s hair.

Explanation:

Efficiency = output/input × 100%

Data

Efficiency = 64%

Energy input = 1000J

Energy output = ?

Efficiency = Energy output/Energy input × 100%

64% = Energy output /1000J × 100%

Divide both sides by 100%

0.64 = Energy output/1000J

Energy output = 0.64 × 1000J

Efficiency output = 640J

Therefore the hair dryer will transfer 640J of energy usefully to Nora’s hair.

6 0

2 years ago

A projectile is launched with an initial velocity of 25 m/s at an angle of 30° above the horizontal. The projectile reaches maxi

Alchen [17]

Answer:

x ≈ 56 m

Explanation:

vertical initial velocity = $v_{0y}(t)$ = 25 m/s* sin(30°)= 12.5 m/s

height = h

$h =v_{0y}t+\frac{at^{2}}{2} \\\\65m = 12.5m/s*t + \frac{9.8m/s^{2}*t^{2}}{2} \\\\t=2.584 s$

t- time is found solving quadratic equation.

horizontal velocity = $v_{0x}=25m/s*cos(30^{o})=21.65 m/s$

Horizontal velocity is constant, so distance $x=v_{0x}*t =21.65 m/s *2.584 s=55.9 = 56 m$

6 0

4 years ago

A submarine is 2.99 ✕ 102 m horizontally from shore and 1.00 ✕ 102 m beneath the surface of the water. A laser beam is sent from

Vsevolod [243]

Answer:

a The diagram of the situation is shown on the first uploaded image

b the angle of incidence beam striking the water is $\theta = 49.63^o$

c the angle of refraction beam striking the water is $r = 59.7^o$

d The angle the refracted beam make with respect to the horizontal is $= 30.3^o$

e The height of the target above sea level is

$h= 125.05m$

Explanation:

From the diagram we see that the angle of the beam striking the water is

$tan \theta = \frac{100}{85}$

$\theta = tan^{-1}(\frac{100}{85} )$

$= 49.63^o$

According to Snell's law

$\mu_{water} *sin(i) = \mu_{air} *sin(r)$

Where $\mu_{water }$ is the refractive index of water = 1.333

$i$ is the angle of incidence

$\mu_{air}$ is the refractive index of air = 1

r is the angle of refraction

Substituting values accordingly

$1.33 * sin (40.37) = 1 * sin(r)$

Making r the subject of the formula

$r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})$

$= 59.7^o$

looking at the diagram we can see that to obtain the angle the refraction beam makes with the horizontal by subtracting the angle refraction from 90°

i.e $90 -59.7 = 30.3$ °

From the diagram we see that the height target above sea level can be obtained by this relation

$tan \theta = \frac{h}{214}\\$

Where h is the is the height

$tan(30.3) = \frac{h}{214}$

$h = 214 * tan (30.3)$

$=125.05m$

4 0

4 years ago