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4 years ago

8

The intensity level is 65 db at a distance 5.00 m from a barking dog. What would be a reasonable estimate for the intensity leve

l if two identical dogs very close to each other were barking? You can ignore any interference effects.

1 answer:

BARSIC [14]4 years ago

7 0

Answer:

68 db

Explanation:

Since now instead of one two dogs are barking simultaneously close to each other, therefore we take n =2.

Ignoring interference effects, the barking of two dogs result in a higher level of intensity which is given by,

β(db)=10×㏒(2)

=3 db

So, a reasonable estimate for the raised Intensity Level is: 65db+3db = 68db

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Part 1) Which metal will cool the fastest?
To answer this question, we should have a look at the formula of the heat flow rate, which says "how fast" a material is able to heat/cool:
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k is thermal conductivity of the material
A the surface where the exchange of heat occurs
$\Delta T$ the variation of temperature
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We see that the heat flow rate $\frac{\Delta Q}{\Delta t}$ is linearly proportional to k, the thermal conductivity of the material. So, the larger k, the fastest the metal will cool.
If we have a look at the thermal conductivity of each metal, we find:
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- Copper: 401 W/(mK)
- Gold: 314 W/(mK)
- Platinum: 69 W/(mK)
Therefore, copper is the material with highest heat flow rate, so the metal which cools fastest.

Part 2) Which sample of copper demonstrates the greatest increase in temperature
To solve this part, we can have a look at how the amount of heat exchanged Q is related to the increase in temperature $\Delta T$ :
$Q=m C_S \Delta T$
where m is the mass and Cs the specific heat of the material. Re-arranging the formula, we get
$\Delta T= \frac{Q}{m C_s}$
therefore, we see that the increase in temperature is inversely proportional to the mass m. This means that the block that will show the largest increase in temperature is the block with the smallest mass, so the correct answer is A) 0.5 kg.

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4 years ago

Which is style of conclusion used in research reports

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A conclusion is, in some ways, like your introduction. You restate your thesis and summarize your main points of evidence for the reader.You can usually do this in one paragraph.

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3 years ago

A 250 kg car has 6875 kg•m/s of momentum. What is it’s velocity?

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Answer:

v = 27.5 m/s

Explanation:

p = m × v

6,875 = 250 × v

250v = 6,875

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3 years ago

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4 years ago

A small wooden block with mass 0.750 kg is suspended from the lower end of a light cord that is 1.72 m long. The block is initia

Ierofanga [76]

Answer:

$v_{0}=319.2 m/s$

Explanation:

We need to use the momentum and energy conservation.

$p_{0}}=p_{f}$

$mv_{0}=(m+M)V_{1}$

Where:

m is the mass of bullet (m=0.01 kg)
M is the mass of the wooden (M=0.75 kg)
v(0) initial velocity of bullet
V(1) is the velocity of the combined object in the moment the bullet hist the block

Conservation of energy.

We have kinetic energy at first and kinetic and potential energy at the end.

$(1/2)(m+M)V_{1}^{2}=(1/2)(m+M)V_{2}^{2}+(m+M)gh$

Here:

V(1) is the velocity of the combined at the initial position
h is the vertical height ( h = 0.800 m)

We can find V(2) using the definition of force at this point:

$\Sigma F=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)$

$T-(m+M)gcos(\theta)=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)$

$cos(\theta) =(L-h)/L=(1.72-0.8)/1.72$

$\theta =57.66$

Now, we can solve the equation to find V(2)

$V_{2}=\sqrt{\frac{R*(T-(m+M)*g*cos(\theta))}{(m+M)}}$

$V_{2}=\sqrt{\frac{1.72*(4.86-(0.01+0.75)*9.81*cos(57.66))}{(0.01+0.75)}}$

$V_{2}=1.40 m/s$

Now we can find V(1) using the conservation of energy equation

$(1/2)V_{1}^{2}=(1/2)V_{2}^{2}+gh$

$V_{1}=\sqrt{V_{2}^{2}+2gh}$

$V_{1}=\sqrt{1.40^{2}+2*9.81*0.8}$

$V_{1}=4.20 m/s$

Finally, using the momentum equation we find v(0)

$v_{0}=\frac{(m+M)V_{1}}{m}$

$v_{0}=\frac{(0.01+0.75)*4.20}{0.01}$

$v_{0}=319.2 m/s$

I hope it helps you!

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4 years ago