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3 years ago

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Sound wave A delivers 2 J of energy in 2 s. Sound wave B delivers 10 J of energy in 5 s. Sound wave C delivers 2 mJ of energy in

1 ms. Rank in order, from largest to smallest, the sound powers of Pa, Pb, Pc of these three waves.

1 answer:

Dmitry [639]3 years ago

8 0

Answer:

In the order largest to smallest

Pc = Pb, Pa

Explanation:

Power = Energy expended/time

Pa = 2/2 = 1Watt

Pb = 10/5 = 2Watt

Pc = 2E-3/1E-3 = 2Watt

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10. Solve the following numerical problems

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Answer:

$\boxed {\boxed {\sf 120 \ Joules}}$

Explanation:

Work is equal to the product of force and distance.

$W=F*d$

The force is 8 Newtons and the distance is 15 meters.

$F= 8 \ N \\d= 15 \ m$

Substitute the values into the formula.

$W= 8 \ N * 15 \ m$

Multiply.

$W= 120 \ N*m$

1 Newton meter is equal to 1 Joule
Our answer of 120 N*m equals 120 J

$W= 120 \ J$

The work done is <u>120 Joules</u>

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How long would it take a leopard, running at an average speed of 20 m/s to travel 500 m?

alexandr1967 [171]

Answer:

25 seconds

Explanation:

500/20

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3 years ago

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr

mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = $5 \ * (M \ _{sun})$

where : $M_{sun} = 1.99*10^{33} \ kg$

Then; we have:

$m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg$

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear $m_{ear} = 0.030 \ kg$

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

$\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R - d) \omega^2$

$\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)$

The net force on the ear farther to the black hole is :

$\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R + d) \omega^2$

$\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)$

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

$T = \frac{3GMm_{ear}d}{R^3}$

$T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}$

$T = 2073.9 N\\T = 2.07 KN$

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

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4 years ago

Find the y-component of this

Alborosie

Answer:

-0.0789 m

Explanation:

Recall that the y-component comes associated with the sin(18.4) through the following trigonometric relationship:

y = 0.250 sin(-18.4) ≈ -0.0789 m

Notice it is negative since it is below the x-axis.

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Two light bulbs have resistances of 400 Ω and 800 ΩThe two light bulbs are connected in series across a 120- V line. Find the cu

Natasha2012 [34]

1) Current in each bulb: 0.1 A

The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

$R_{eq}=R_1 + R_2 = 400 \Omega + 800 \Omega=1200 \Omega$

And so, the current through the circuit is (using Ohm's law):

$I=\frac{V}{R_{eq}}=\frac{120 V}{1200 \Omega}=0.1 A$

And since the two bulbs are connected in series, the current through each bulb is the same.

2) 4 W and 8 W

The power dissipated by each bulb is given by the formula:

$P=I^2 R$

where I is the current and R is the resistance.

For the first bulb:

$P_1 = (0.1 A)^2 (400 \Omega)=4 W$

For the second bulb:

$P_1 = (0.1 A)^2 (800 \Omega)=8 W$

3) 12 W

The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:

$P_{tot} = P_1 + P_2 = 4 W + 8 W=12 W$

3 0

3 years ago