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2 years ago

A wave is traveling at a speed of 15 m/s and it's wavelength is 5 m. Calculate the waves frequency

Physics

1 answer:

Anvisha [2.4K]2 years ago

3 0

Answer:

The frequency of this wave is $3\; \rm Hz$ .

Explanation:

The frequency $f$ of a wave is the number of wavelengths that this wave covers in unit time (typically a second.)

The wave in this question travels at $v = 15\; \rm m \cdot s^{-1}$ . In other words, this wave covers $15\; \rm m$ in unit time (a second.) How many wavelengths $\lambda$ would that $15\; \rm m\;$ correspond to?

The question states that the wavelength of this wave is $\lambda = 5\; \rm m$ . Therefore, there would be $15 / 5 = 3$ wavelengths in the $15\; \rm m$ span that this wave covered in the unit time of one second ( $1\; \rm s$ .) Hence, the frequency of this wave would be $3\; \rm s^{-1}$ (three per second,) which is equivalent to $3\; \rm Hz$ (three Hertzs.)

In general, the frequency $f$ of a wave with speed $v$ and wavelength $\lambda$ would be:

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

For the wave in this question:

$\begin{aligned}f &= \frac{v}{\lambda} \\ &= \frac{15\; \rm m \cdot s^{-1}}{3\; \rm s} = 3\; \rm s^{-1} = 3\; \rm Hz\end{aligned}$ .

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In a series lrc circuit, the frequency at which the circuit is at resonance is f0. If you double the resistance, the inductance,

taurus [48]

When you double capacitance and inductance, the new resonance frequency becomes f/2.

Resonance frequency:

The resonance frequency of RLC series circuit, is the frequency at which the capacity reactance is equal to inductive reactance.

It can also be defined as the natural frequency of an object where it tends to vibrate at a higher amplitude.

Xc = Xl

which gives the value for resonance frequency:

$f = \frac{1}{2\pi \sqrt{LC} }$

where;

f is the resonance frequency

L is the inductance

C is the capacitance

When you double capacitance and inductance, the new resonance frequency becomes;

$f' = \frac{1}{2\pi \sqrt{2L2C} }$

$f' = \frac{1}{2\pi \sqrt{4LC} }$

$f' = \frac{1}{\pi \sqrt{LC} }\frac{1}{2}$

$f' = \frac{1}{2} f$

Thus from above,

When you double capacitance and inductance, the new resonance frequency becomes f/2.

Learn more about resonance frequency here:

<u>brainly.com/question/13040523</u>

#SPJ4

6 0

1 year ago

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .