All categories

3 years ago

Why does the frequency of a wave increase as the wavelength decreases

2 answers:

5 0

Because their PRODUCT is always the same number ... the wave speed. So if one of them changes, the other one has to change in the opposite direction, in order to keep their product the same.

olchik [2.2K]3 years ago

3 0

Explanation:

We know that the number of complete waves formed in 1 sec time is frequency and the distance between two consecutive crests or troughs is wavelength. And we have the formula that

Velocity = wavelength * frequency

or, frequency = velocity / wavelength

Here we can see frequency is directly proportional to velocity and indirectly proportional to wavelength.

So as the wavelength increases frequency decreases and as the wavelength decreases frequency increases.

Hope you understood

You might be interested in

A wire is made from a material having a temperature coefficient of resistivity of 0.0003125 (°C^-1). In an experiment, we mainta

beks73 [17]

Answer:

The power decreases by 36%

Explanation:

Given:

At 20° C

Power, P₀ = 300 W

Potential difference, V = 150 volts

Now, power is given as

P = V²/R

where, R is the resistance

on substituting the values, we get

300 = 150²/R₀

R₀ = 75 Ω

Now, the variation of resistance with temperature is given as

R = R₀[1 + α(T - T₀)]

where, α is the temperature coefficient of resistivity = 0.0003125 (°C⁻¹)

now, at

T₀ = 20° C

R₀ = 75 Ω

for

T = 1820° C

we have

R = R₀[1 + α(T - T₀)]

substituting the values

we get

R = 75×[1 + 0.0003125 × (1820 - 20)]

R = 117.18 Ω

Now using the formula for power

We have,

P = V²/R

P = 150²/117.18 = 192 W

Therefore, the percentage change will be

= $\frac{P-P_0}{P_0}\times 100$

on substituting the values , we get

= $\frac{192-300}{300}\times 100$

= -36%

here, negative sign depicts the decrease in power

3 0

2 years ago

Read 2 more answers

A 0.10 g honeybee acquires a charge of +23 pC while flying.

kari74 [83]

Answer:

a) $\frac{F}{w} =2.347\times 10^{-6}\ N$

b) $E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, $m=10^{-4}\ kg$

charge acquired by the bee, $q_2=23\times 10^{-12}\ C$

Electrical field near the earth surface, $E=100\ N.C^{-1}$

Now the electric force on the bee:

we know:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}$

$F=E.q_2$

$F=100\times 23\times 10^{-12}$

$F=23\times 10^{-10}\ N$

The weight of the bee:

$w=m.g$

$w=10^{-4}\times 9.8$

$w=9.8\times10^{-4}\ N$

Therefore the ratio :

$\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}$

$\frac{F}{w} =2.347\times 10^{-6}\ N$

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

$F=9.8\times10^{-4}\ N$

$E.q_2=9.8\times10^{-4}\ N$

$E\times 23\times 10^{-12}=9.8\times10^{-4}\ N$

$E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

3 0

3 years ago

A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at

stellarik [79]

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

$\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft$

Also, at any instant t

$\Rightarrow l^2=x^2+y^2$

differentiate w.r.t.

$\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s$

5 0

3 years ago

The graph shows the range of frequencies commonly heard by some animals.

Vsevolod [243]

Concept:

Frequency- It is defined as the number of oscillations occur in one second.

Its SI unit is Hertz (Hz)

Given: Produced sound vibrations is 18,500 cycles in 0.75 seconds

∵ In 0.75 second, produced sound has oscillations = 18,500 cycles

∴ In 1.0 second, produced sound has oscillations = (18,500 ÷ 0.75) Hz

The frequency of the sound will be ≈ 24,667 Hz

From the study of the given graph, only the animals (c) Cats, (b) Moths and (a) Bats can hear the produced sound because their upper audible frequency range is greater than 24,667 Hz.

3 0

3 years ago

Read 2 more answers

a child riding a bicycle at 15 meters per second accelerates at -3,0 meters per second? for 4.0 seconds. What is the child's spe

Vika [28.1K]

Answer:

Explanation:

Given data

initital velocity u=15m/s

deceleration a=3m/s^2

time t= 4 seconds

final velocity v= ?

Applying the expression

v=u+at------1

substituting our data into the expression we have

v=15+3*4

v=15+12

v=27m/s

The velocity after 4 seconds is 27m/s

5 0

2 years ago