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2 years ago

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The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 km. Assume the

acoustic output of a howler to be uniform in all directions and that the threshold of hearing is 1.0 × 10-12 W/m2. A juvenile howler monkey has an acoustic output of 63 ȝW. What is the ratio of the acoustic intensity produced by the juvenile howler to the reference intensity I0, at a distance of 210 m?

1 answer:

exis [7]2 years ago

7 0

Answer:

113.7

Explanation:

maximum distance (s) = 8.9 km

reference intensity (I0) = 1 x 10^{-12} W/m^{2}

power of a juvenile howler monkey (p) = 63 x 10^{-6} W

distance (r) = 210 m

intensity (I) = power/area

where we assume the area of a sphere due to the uniformity of the output in all directions

area = 4π $r^{2}$ = 4π x $210^{2}$ = 554,176.9 m^{2}

intensity (I) = $\frac{63 x 10^{-6} }{554,176.9} = 113.7 x 10^{-12}$

therefore the desired ratio I/I0 = $\frac{113.7 x 10^{-12}}{1 x 10^{-12}}$ = 113.7

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A rifle, which has a mass of 5.50 kg., is used to fire a bullet, which has a massof m = 65.0 grams., at a "ballistics pendulum".

Alex787 [66]

Answer:

Part a)

$U = 13 J$

Part b)

$v = 2.28 m/s$

Part c)

$v = 177.66 m/s$

Part d)

$W = 1012.7 J$

Part e)

$v = 2.1 m/s$

Part f)

$E = 1037.2 J$

Explanation:

Part a)

As we know that the maximum angle deflected by the pendulum is

$\theta = 38^o$

so the maximum height reached by the pendulum is given as

$h = L(1 - cos\theta)$

so we will have

$h = L(1 - cos38)$

$h = 1.25(1 - cos38)$

$h = 0.265 m$

now gravitational potential energy of the pendulum is given as

$U = mgh$

$U = 5(9.81)(0.265)$

$U = 13 J$

Part b)

As we know that there is no energy loss while moving upwards after being stuck

so here we can use mechanical energy conservation law

so we have

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.81)(0.265)}$

$v = 2.28 m/s$

Part c)

now by momentum conservation we can say

$mv = (M + m) v_f$

$0.065 v = (5 + 0.065)2.28$

$v = 177.66 m/s$

Part d)

Work done by the bullet is equal to the change in kinetic energy of the system

so we have

$W = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)v_f^2$

$W = \frac{1}{2}(0.065)(177.66)^2 - \frac{1}{2}(5 + 0.065)2.28^2$

$W = 1012.7 J$

Part e)

recoil speed of the gun can be calculated by momentum conservation

so we will have

$0 = mv_1 + Mv_2$

$0 = 0.065(177.6) + 5.50 v$

$v = 2.1 m/s$

Part f)

Total energy released in the process of shooting of gun

$E = \frac{1}{2}Mv^2 + \frac{1}{2}mv_1^2$

$E = \frac{1}{2}(5.50)(2.1^2) + \frac{1}{2}(0.065)(177.6^2)$

$E = 1037.2 J$

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2 years ago

The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume tha

gregori [183]

The answers are:

a) $Work=125,923.61J$

b) $Power=1.46watt$

Why?

It seems that you forgot to write the questions of the problem, however, in order to help you, I will try to complete it.

The questions are:

a) How much work does the heart do in a day?

b) What is its power output in watts?

So, solving we have:

We need to convert from liter to cubic meters in order to use the given information, so:

$1L=0.001m^{3}\\\\7500L*\frac{0.001m^{3} }{1L}=7.5m^{3}$

Also, we need to find the mass given the density of the blood.

$1050}\frac{kg}{m^{3}}*7.5m^{3}=7875kg$

Now, calculating how much work does the heart do in a day, we have:

$Work=Fd=mgh\\\\Work=7875kg*9.81\frac{m}{s^{2}}*1.63m=125,923.61J$

Then, calculating what is the power output and its horsepower, we have:

$Power=\frac{Work}{time}\\\\Power=\frac{125,923.61J}{86,400s}=1.46watt$

Have a nice day!

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A Wooden block has a mass of 0.200kg, a specific heat of 710 J (kg times degrees Celsius and is at a temperature of 20.0 degrees

olchik [2.2K]

Answer:

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Explanation:

q = mCΔT

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T = 35°C

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2 years ago

Engineers who design battery-operated devices such as cell phones and MP3 players try to make them as efficient as possible. An

Svetradugi [14.3K]

efficiency= [useful energy transferred ÷ total energy supply]×100%

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=55%

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2 years ago