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4 years ago

A wave on a small lake has an amplitude of 0.1. What happens to the energy of the wave by doubling its height to 0.2 meters

Physics

1 answer:

svetoff [14.1K]4 years ago

3 0

As we know that energy of wave depends on amplitude as per following relation

$E = kA^2$

here we know that

k = constant

A = amplitude of wave

here we know that amplitude is doubled

so here as per above formula the energy of wave will increase by 4 times

as we know that

$E = k(2A)^2$

so energy is 4 times of initial value

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3 years ago

A woman (mass= 50.5 kg) jumps off of the ground, and comes back down to the ground at a velocity of -8.4 m/s.

Blizzard [7]

Answer:

Approximately $1.6\times 10^{3}\; \rm N$ .

Explanation:

By the Impulse-Momentum Theorem, the change in this woman's momentum will be equal to the impulse that is applied to her.

The momentum $p$ of an object is equal to the product of its mass $m$ and velocity $v$ . That is: $p = m \cdot v$ .

Let $v(\text{before})$ and $v(\text{after})$ represent the velocity of the woman before and after the landing. Let $m$ represent the woman's mass.

The woman's momentum before the landing would be $m \cdot v(\text{before})$ .
The woman's momentum after the landing would be $m \cdot v(\text{after})$ .

Therefore, the change in this woman's momentum would be:

$\begin{aligned}& \Delta p \\ & = p(\text{after}) - p(\text{before}) \\ &= m \cdot (v(\text{after})- v(\text{before}))\end{aligned}$ .

On the other hand, impulse is equal to force multiplied by the duration of the force. Let $F$ represent the average force on the woman. The impulse on her during the landing would be $F \cdot t$ .

Apply the Impulse-Momentum Theorem.

Impulse: $F\cdot t$ .
Change in momentum: $m \cdot (v(\text{after})- v(\text{before}))$ .

Impulse is equal to the change in momentum:

$F \cdot t = m \cdot (v(\text{after})- v(\text{before}))$ .

After landing, the woman comes to a stop. Her velocity would become zero. Therefore, $v(\text{after}) = 0\; \rm m \cdot s^{-1}$ .

$\begin{aligned}F &= \displaystyle \frac{m \cdot (v(\text{after})- v(\text{before}))}{t} \\ &= \frac{50.5\; \text{kg} \times \left(0 \; \mathrm{m \cdot s^{-1}}- 8.4\; \mathrm{m \cdot s^{-1}}\right)}{0.27\; \rm s} \\ &\approx 1.6 \times 10^{3}\; \rm N\end{aligned}$ .