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erastovalidia [21]

3 years ago

7

A striped billiard ball rolls and strikes a motionless solid billiard ball. The solid ball then begins to roll. Which best descr

ibes the change in the magnitudes of momentum for the balls?

1 answer:

professor190 [17]3 years ago

4 0

This situation can be best explained by the conservation of momentum. The momentum of the rolling ball is passed on to the motionless solid ball. After collision, the then rolling ball slows to a stop, while the then motionless solid ball moves with the same magnitude of momentum as that of the striped billiard ball.

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The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan

ss7ja [257]

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

$I = \frac{P}{A}$

The area of a sphere is given by

$A = 4\pi r^2$

So replacing we have to

$I = \frac{P}{4\pi r^2}$

Since the question tells us to find the proportion when

$r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}$

So considering the two intensities we have to

$I_1 = \frac{P_1}{4\pi r_1^2}$

$I_2 = \frac{P_2}{4\pi r_2^2}$

The ratio between the two intensities would be

$\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}$

The power does not change therefore it remains constant, which allows summarizing the expression to

$\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2$

Re-arrange to find $I_2$

$I_2 = I_1 (\frac{r_1}{r_2})^2$

$I_2 = 100*(\frac{1}{5})^2$

$I_2 = 4W/m^2$

Therefore the intensity at five times this distance from the source is $4W/m^2$

3 0

3 years ago

We wrap a light, nonstretching cable around a 8.00 kg solid cylinder with diameter of 30.0 cm. The cylinder rotates with negligi

antoniya [11.8K]

Answer:

h = 16.67m

Explanation:

If the kinetic energy of the cylinder is 510J:

$Kc=510=1/2*Ic*\omega c^2$

$\omega c=\sqrt{510*2/Ic}$

Where the inertia is given by:

$Ic=1/2*m_c*R_c^2=1/2*(8)*(0.15)^2=0.0225kg.m^2$

Replacing this value:

$\omega c=106.46rad/s$

Speed of the block will therefore be:

$V_b=\omega_c*R_c=106.46*0.15=15.969m/s$

By conservation of energy:

Eo = Ef

Eo = 0

$Ef = 510+1/2*m_b*V_b^2-m_b*g*h$

So,

$0 = 510+1/2*m_b*V_b^2-m_b*g*h$

Solving for h we get:

h=16.67m

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Virty [35]

Answer:

I only speak English

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Answer:

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Explanation:

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2 years ago