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2 years ago

3

An empty spice bottle has an inner volume of 1.31 ´ 10-4 m3. It has a mass of 112 g when filled with air, and it displaces 1.63

´ 10-4 m3 of water when fully submerged. What fraction of the total volume of the bottle will be beneath the surface when it is placed in a tank of water?

1 answer:

Stells [14]2 years ago

3 0

Answer:

The fraction of the total volume of the bottle will be beneath the surface is 0.687

Explanation:

The immersed volume is equal to:

$mg=V_{imm} *\rho _{water} *g\\V_{imm}=\frac{m}{\rho _{water}}$

Where

m = 112 g = 0.112 kg

ρwater = 1000 kg/m³

Replacing:

$V_{imm} =\frac{0.112}{1000} =1.12x10^{-4} m^{3}$

The fraction of the total volume of the bottle will be beneath the surface is:

$fraction=\frac{1.12x10^{-4} }{1.63x10^{-4} } =0.687$

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vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

$Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}$

$Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V$

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

$R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200$

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.

if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

$Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V$

$Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V$

$R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140$

$R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260$

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$V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}$

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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:

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

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Have a nice day!

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A 145-g baseball is thrown so that it acquires a speed of 25 m/s. What was the net work done on the ball to make it reach this s

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When the ball has left your hand and is flying on its own, its kinetic energy is

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KE = (0.0725 kg) (625 m²/s²)

<em>KE = 45.3 Joules</em>

If the baseball doesn't have rocket engines on it, or a hamster inside running on a treadmill that turns a propeller on the outside, then there's only one other place where that kinetic energy could come from: It MUST have come from the hand that threw the ball. The hand would have needed to do <em>45.3 J</em> of work on the ball before releasing it.

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yKpoI14uk [10]

Answer:

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Explanation:

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This is newton’s second laws.

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In mathematically,

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r = distance between both objects

Hence, The newton’s second law is F=ma

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