What important example, shown in the picture, is given regarding a real-life

You are sitting 3 m away from you friend who is watching a cartoon on his phone. How will the sound itensity change if your frie

Zinaida [17]

Answer:

Decreases by $3.6 \times 10^{-3}$ times

Explanation:

The intensity of a sound is defined as the energy of the sound that is flowing in an unit time through the unit area which is in the direction that is perpendicular to the direction of the sound waves movement.

The intensity of energy is described by the inverse square law. It states that the intensity varies inversely with the distance square of the distance.

In other words, the sound intensity decreases as inversely proportional to the squared of the distance. i.e. $\frac{1}{r^2}$

In the context when the distance was 3 m, the intensity of the sound was = $\frac{1}{9}$

But when the distance became 6 cm or 0.06 m, the sound intensity decreases by = $\frac{1}{0.06^2}$

= $3.6 \times 10^{-3}$ times

3 0

3 years ago

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$