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3 years ago

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If a topographic map included a 6,000 ft. mountain next to an area of low hills, which would best describe the contour lines on

the map? a. The contour lines would be dark blue. b. The contour lines around the mountain would be very close together. c. The contour lines would cross near the top of the mountain. d. The contour lines around the rolling hills would be very close together.

2 answers:

VMariaS [17]3 years ago

8 0

The answer to the given question above would be option B. If a topographic map included a 6,000 ft. mountain next to an area of low hills, the statement that best describe the contour lines on the map is this: <span>The contour lines around the mountain would be very close together. Hope this helps.</span>

boyakko [2]3 years ago

7 0

The answer to your question is B. 100%

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Answer:

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3 years ago

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

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Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

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

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

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$P = 6.1*10^{-9}J/s$

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3 years ago

Please help me out !!

blagie [28]

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2 years ago

How long does it take Johnny to reach a speed of 80 m/s in his car while accelerating at 10m/s per second?

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Answer:

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3 years ago

In an elastic collision, a 300 kg bumper car collides directly from behind with a second, identical bumper car that is traveling

evablogger [386]

Answer:

If we had:

$v_{1i}=5.3m/s$

$v_{2i}=5.9m/s$

We will have:

$v_{1f}=5.9m/s$

$v_{2f}=5.3m/s$

Explanation:

In an elastic collision both linear momentum and kinetic energy are conserved, so we will have:

$p_i=p_f$

$K_i=K_f$

We will call our bumpers 1 and 2.

For the momentum equation we know that:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

Since all the masses are the same (300kg), they cancel out:

$v_{1i}+v_{2i}=v_{1f}+v_{2f}$

For the kinetic energy equation we know that:

$\frac{m_1v_{1i}^2}{2}+\frac{m_2v_{2i}^2}{2}=\frac{m_1v_{1f}^2}{2}+\frac{m_2v_{2f}^2}{2}$

Since all the masses are the same (300kg), they cancel out (and also the 2 dividing):

$v_{1i}^2+v_{2i}^2=v_{1f}^2+v_{2f}^2$

We then must solve this system:

$v_{1i}+v_{2i}=v_{1f}+v_{2f}$

$v_{1i}^2+v_{2i}^2=v_{1f}^2+v_{2f}^2$

Which we will rewrite as:

$v_{1i}-v_{1f}=v_{2f}-v_{2i}$

$v_{1i}^2-v_{1f}^2=v_{2f}^2-v_{2i}^2$

The last of these equations can be written as:

$(v_{1i}+v_{1f})(v_{1i}-v_{1f})=(v_{2f}+v_{2i})(v_{2f}-v_{2i})$

But we know that $v_{1i}-v_{1f}=v_{2f}-v_{2i}$ , so those cancel out:

$v_{1i}+v_{1f}=v_{2f}+v_{2i}$

So we can write:

$v_{1i}-v_{1f}+v_{2i}=v_{2f}$

$v_{1i}+v_{1f}-v_{2i}=v_{2f}$

Which means:

$v_{1i}-v_{1f}+v_{2i}=v_{1i}+v_{1f}-v_{2i}$

Which solving for the final velocity leaves us with:

$v_{2i}+v_{2i}=+v_{1f}+v_{1f}$

$v_{1f}=v_{2i}$

Grabbing any equation that relates both final velocities easily, for example $v_{1i}-v_{1f}+v_{2i}=v_{2f}$ , we obtain:

$v_{2f}=v_{1i}-v_{1f}+v_{2i}=v_{1i}-v_{1f}+v_{1f}=v_{1i}$

So we conclude that the bumpers have just exchanged velocities (something sometimes seen in billiards for example):

$v_{1f}=v_{2i}=5.9m/s$

$v_{2f}=v_{1i}=5.3m/s$

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3 years ago