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2 years ago

How are wavelength and frequency related for a wave moving at a constant speed?

1 answer:

soldier1979 [14.2K]2 years ago

6 0

Mathematically, to compute for the wavelength, λ, we have

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

where v is the velocity and f is the frequency.
As we can see here, given that velocity is constant, wavelength increases as frequency decreases. This shows that wavelength is inversely proportional to frequency.

Answer: Wavelength and frequency are inversely related to each other.

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Please help me with this physics prooblem

zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

$x=v_0\cos20.0^\circ t+\dfrac12a_xt^2$

$y=v_0\sin20.0^\circ t+\dfrac12a_yt^2$

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

$x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ$

$y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ$

So we have enough information to solve for the components of the acceleration vector, $a_x$ and $a_y$ :

$x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}$

$y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}$

The acceleration vector then has direction $\theta$ where

$\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ$

5 0

2 years ago

A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of

posledela

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

T2 = 23 °C = 273 + 23 = 300 K

Let the volume of bubble at the surface of the lake is V2.

Density of water, d = 1000 kg/m^3

Po = atmospheric pressure = 10^5 N/m^2

P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

By substituting the values, we get

$\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}$

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

6 0

3 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

2 years ago

A 250 - ω resistor is connected in series with a 4.80 - μf capacitor. the voltage across the capacitor is vc=(7.60v)⋅sin[(120rad

galben [10]

<span>553 ohms The Capacitive reactance of a capacitor is dependent upon the frequency. The lower the frequency, the higher the reactance, the higher the frequency, the lower the reactance. The equation is Xc = 1/(2*pi*f*C) where Xc = Reactance in ohms pi = 3.1415926535..... f = frequency in hertz. C = capacitance in farads. I'm assuming that the voltage and resistor mentioned in the question are for later parts that are not mentioned in this question. Reason is that they have no effect on the reactance, but would have an effect if a question about current draw is made in a later part. With that said, let's calculate the reactance. The 120 rad/s frequency is better known as 60 Hz. Substitute known values into the formula. Xc = 1/(2*pi* 60 * 0.00000480) Xc = 1/0.001809557 Xc = 552.6213302 Rounding to 3 significant figures gives 553 ohms.</span>

6 0

3 years ago

Read 2 more answers

What happens when white light shines onto a red filter?

Gnesinka [82]

The dye in between the two pieces of glass in the filter absorbs all the other colors in the white light except the red, and the red is the only light left to come out the other side. That's why the filter usually looks kind of red. It's also a big part of the reason why we call it a "red filter".

6 0

3 years ago