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2 years ago

How can you double the frequency of a wave if you have control over both the wavelength and the wave velocity?

Physics

2 answers:

iren2701 [21]2 years ago

6 0

Answer:

Either double the speed or reduce the wavelength to half

Explanation:

We know that velocity of the wave is given by velocity=wavelength ×frequency

$v=\lambda f$

So $f=\frac{velocity}{wavelength}=\frac{v}{\lambda }$

As in the question it is given that we have control on both velocity and wavelength

So we have to double the the frequency then we can double the speed or reduce the wavelength to half

sammy [17]2 years ago

5 0

Answer:

Change the wavelength to half while keeping velocity constant or change the wave velocity to 2 times while keeping wavelength constant.

Explanation:

The frequency of a wave can be defined as the rate of change of wave speed with respect to the wavelength of the wave.

Mathematically,

$f=\frac{v}{\lambda}$

Here, v is the velocity, lambda is the wavelength, and f is the frequency of the wave.

If the observer want to double the frequency of the wave, then he should increase the wave velocity two times and take wavelength constant for this case or either he should half the wavelength of the wave to take the velocity of the wave constant.

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Plastic is what they are made of

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3 years ago

I can't seem to get the right angular acceleration and also not sure how to do part b. Help will be much appreciated.

Tcecarenko [31]

Answer:It’s 5 I believe

Explanation: it says to round to the nearest thousandths, so it’ll be 5.

8 0

2 years ago

A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.

matrenka [14]

Answer:

Part a)

$T = 42 N$

Part b)

$v_f = 11.8 m/s$

Part c)

$t = 1.7 s$

Part d)

$F = 159.7 N$

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

$mg - T = ma$

for uniform cylinder we will have

$TR = I\alpha$

so we have

$T = \frac{1}{2}MR^2(\frac{a}{R^2})$

$T = \frac{1}{2}Ma$

now we have

$mg = (\frac{M}{2} + m)a$

$a = \frac{mg}{(\frac{M}{2} + m)}$

$a = \frac{15 \times 9.81}{(6 + 15)}$

$a = 7 m/s^2$

now we have

$T = \frac{12 \times 7}{2}$

$T = 42 N$

Part b)

speed of the bucket can be found using kinematics

so we have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(7)(10)$

$v_f = 11.8 m/s$

Part c)

now in order to find the time of fall we can use another equation

$v_f - v_i = at$

$11.8 - 0 = 7 t$

$t = 1.7 s$

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

$F = T + Mg$

$F = 42 + (12 \times 9.81)$

$F = 159.7 N$

5 0

3 years ago

A 2 kg blue car is moving 6 m/s to the right and collides with a 3 kg red car that is moving 2 m/s to the left. The cars collide

snow_lady [41]

Answer:

Their velocity after the collision is 1.2 m/s, to the right.

Explanation:

Given;

mass of the blue car, m₁ = 2 kg

initial velocity of the blue car, u₁ = 6 m/s

mass of the red car, m₂ = 3 kg

initial velocity of the red car, u₂ = 2 m/s

let the blue car moving to the right be in positive direction

also, let the red car moving to the left be in negative direction

Apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ - m₂u₂ = v(m₁ + m₂)

where;

v is their velocity after the collision

(2 x 6) - (3 x 2) = v(2 + 3)

12 - 6 = 5v

6 = 5v

v = 6/5

v = 1.2 m/s, to the right

Therefore, their velocity after the collision is 1.2 m/s, to the right.

7 0

2 years ago

A spherical submersible 1.84 m in radius, armed with multiple cameras, descends under water in a region of the Atlantic Ocean kn

Drupady [299]

Answer:

23932242.5 Pa

Explanation:

$P_a$ = Atmospheric pressure = $1.013\times 10^5\ Pa$

$P_w$ = Pressure of seawater

$\rho$ = Density of sea water = $1.025\times 10^3\ kg/m^3$

h = Depth of shipwreck = $2.37\times 10^3\ m$

g = Acceleration due to gravity = 9.81 m/s²

The absolute pressure is given by

$P_{ab}=P_a+P_w\\\Rightarrow P_{ab}=1.013\times 10^5+1.025\times 10^3\times 9.81\times 2.37\times 10^3\\\Rightarrow P_{ab}=23932242.5\ Pa$

The absolute pressure at the depth of the shipwreck is 23932242.5 Pa

5 0

3 years ago