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3 years ago

What happens to the wavelength of a wave if the frequency quadruples, but the wave is in the same medium?

Physics

1 answer:

Otrada [13]3 years ago

7 0

Answer:

I think C? I'm not sure totally though...

Explanation:

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A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 Î¼C on each plate. The plates have a

butalik [34]

Answer:

The potential difference between the plates is 596.2 volts.

Explanation:

Given that,

Capacitance $C=260\ pF$

Charge $q=0.155\ \mu\ C$

Separation of plates = 0.313 mm

We need to calculate the potential difference between the plates

Using formula of potential difference

$V= \dfrac{Q}{C}$

Where, Q = charge

C = capacitance

Put the value into the formula

$V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}$

$V=596.2\ volts$

Hence,The potential difference between the plates is 596.2 volts.

7 0

3 years ago

Two vehicles, a car and a truck, leave an intersection at the same time. the car heads east at an average speed of 50 miles pe

olya-2409 [2.1K]

The car heads east at an average speed of 50 miles per hour from the intersection point towards East. The truck heads east at an average speed of 60 miles per hour from the intersection point towards South.

The distance of car from the intersection point after t hours is $50t$ .

The distance of truck from the intersection point after t hours is $60t$ .

Since these distances are perpendicular to each other, distance apart d (in miles) at the end of t hours is

$d=\sqrt{(50t)^2+(60t)^2} \\ d=10\sqrt{61} t\\ d=78.1t$

Thus the distance apart is $d=78.1t \;miles$

5 0

3 years ago

Answer fast !!!!!!! pleaseee

wariber [46]

Answer: Magnetic Fields

Explanation:

5 0

2 years ago

Read 2 more answers

Una patinadora de 50 kg parte del reposo y después de recorrer 3k alcanza una velocidad de 15 m/s. ¿Qué fuerza neta experimenta

e-lub [12.9K]

Answer:

$F_{net} = 1.875\,N$

Explanation:

Asúmase que la patinadora experimenta una aceleración constante. La fuerza neta experimentada por la patinadora:

$F_{net} = (50\,kg)\cdot \left[\frac{\left(15\,\frac{m}{s}\right)^{2}-\left(0\,\frac{m}{s}\right)^{2} }{2\cdot (3000\,m)} \right]$

$F_{net} = 1.875\,N$

6 0

3 years ago

A van is traveling with an initial velocity of 12 m/s. The driver takes a time of 45 seconds to speed up to a velocity of 20 m/s

Rufina [12.5K]

Initial velocity=u=12m/s
Final velocity=v=20m/s
Time=t=45s

$\\ \rm\hookrightarrow Acceleration=\dfrac{v-u}{t}$

$\\ \rm\hookrightarrow Acceleration=\dfrac{20-12}{45}$

$\\ \rm\hookrightarrow Acceleration=\dfrac{8}{45}$

$\\ \rm\hookrightarrow Acceleration=0.1m/s^2$

Now

Distance=s

$\\ \rm\hookrightarrow v^2-u^2=2as$

$\\ \rm\hookrightarrow (20)^2-12^2=2(0.1)s$

$\\ \rm\hookrightarrow 400-144=0.2s$

$\\ \rm\hookrightarrow 256=0.2s$

$\\ \rm\hookrightarrow s=\dfrac{256}{0.2}$

$\\ \rm\hookrightarrow s=1280m$

4 0

3 years ago