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2 years ago

How are wave properties and energy related? EXPLAIN

Physics

1 answer:

dexar [7]2 years ago

6 0

Answer:

The higher the amplitude, the higher the energy. To summarise, waves carry energy. The amount of energy they carry is related to their frequency and their amplitude. The higher the frequency, the more energy, and the higher the amplitude, the more energy.

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Can you help me on this please

Vikentia [17]

Answer: the water level would rise since the pebble displaces minimal water compared to the boat.

Explanation:..........

8 0

3 years ago

The half-life of plutonium 239 is 24,200 years. Assume that the decay rate is proportional to the amount. Determine the amount o

kotykmax [81]

Answer:

time taken is equal to 14,156 years

Explanation:

we know,

$Y=Ae^{-kt}$

at t = 0

Y(0) = A

given that half life of plutonium 239 = 24,200

$\dfrac{A}{2}=Ae^{-kt}\\0.5=e^{-kt}\\k\times 24200 = ln(2)\\k = \dfrac{ ln(2)}{24200}$

$Y=Ae^{-kt}$

$\frac{3}{2} = e^{-kt}\\ln(1.5)=-\dfrac{ ln(2)}{24200}\times t\\t=-\dfrac{ln(1.5)\times 24200}{ ln(2)}\\t=14,156 \ years$

hence time taken is equal to 14,156 years

5 0

3 years ago

You set a tuning fork into vibration at a frequency of 723 Hz and then drop it off the roof of the Physics building where the ac

zaharov [31]

Answer:

Explanation:

Given

Original Frequency $f=723\ Hz$

apparent Frequency $f'=697\ Hz$

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

$f'=f\cdot \frac{v-v_o}{v+v_s}$

where

$f'=$ apparent frequency

v=velocity of sound in the given media

$v_s=$ velocity of source

$v_0=$ velocity of observer

here $v_0=0$

$697=723\cdot (\frac{343-0}{343+v_s})$

$v_s=(\frac{f}{f'}-1)v$

$v_s=(\frac{723}{697}-1)\cdot 343$

$v_s=12.79\approx 12.8\ m/s$

i.e.fork acquired a velocity of $12.8 m/s$

distance traveled by fork is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v_s^2-0=2\times 9.8\times s$

$s=\frac{12.8^2}{2\times 9.8}$

$s=8.35\ m$

5 0

3 years ago

What is the final position of an object that travel 5 meters/ second for 10 seconds?

Naddik [55]

Answer:

$velocity = \frac{displacement}{time} \\ displacement = velocity \times time \\ = 5 \times 10 = 50meter \\ thank \: you$

4 0

3 years ago

A 0.51-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 m to 0.23 m (relative to it

den301095 [7]

Answer:

487.23 N/m

Explanation:

Given:

mass of metal sphere 'm'= 0.51kg

the spring stretches from 0.12 m to 0.23 m. Therefore,

$s_1}$ = 0.12m and $s_{2}$ = 0.23m

the speed of the sphere decreases from 6.7 to 3.3 m/s. Therefore,

$v_{1}$ = 6.7m/s and $v_2}$ =3.3m/s

In order to find spring constant, we apply law of conservation of energy. i.e

The change of the kinetic energy of sphere is equal to the change of potential energy of the spring.

So, Δ $E_{k} =$ Δ $E_{v}$

where,

Δ $E_{k} =$ 1/2 m ( $v_{1}$ - $v_2}$ )²

Δ $E_{v}$ = 1/2 k ( $s_1}$ - $s_{2}$ )²

1/2 m ( $v_{1}$ - $v_2}$ )² = 1/2 k ( $s_{2}$ - $s_1}$ )²

k= m [ ( $v_{1}$ - $v_2}$ )²/( $s_{2}$ - $s_1}$ )²

k= 0.51 [(6.7-3.3)²/ (0.23-0.12)²]

k= 487.23 N/m

Thus, the spring constant of the spring is 487.23 N/m

5 0

3 years ago