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Readme [11.4K]
3 years ago
6

A copper rod 81 cm long is used to poke a fire. the hot end of the rod is maintained at 105° c and the cool end has a constant t

emperature of 21°
c. what is the temperature of the rod 23 cm from the cool end
Physics
1 answer:
Ugo [173]3 years ago
7 0
The temperature inside the copper rod varies linearly with the distance from the hot end of the rod. This means that we can find the temperature at 23 cm (let's call it 'point A') from the cool end by solving a linear proportion.

The temperature difference between the two ends of the rod is
\Delta T = 105^{\circ}-21^{\circ} = 84^{\circ}
and this corresponds to a length of 81 cm. Therefore, we can write:
84^{\circ}:81~cm = x:23~cm
from which we find
x=23.8~^{\circ}
This is not the final answer actually; this is the temperature difference between the cool end and point A. So, the temperature at point A is
T_A = 21^{\circ}+x=44.8^{\circ}
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a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

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So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

cos \theta = \frac {V_{river}}{V_{boat}}

When solving for theta, we get that:

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so now we can substitute the corresponding values:

\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})

Which yields:

\theta = 60^{o}

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

V_{y}=5.60km/hr*cos(210^{o})

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part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

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t_{us}=\frac{2.80km}{2,80km/hr}=1hr

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t=1.333hr

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f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

t=\frac{d}{v_{boat}}

t=\frac{5.60km}{5.60km/hr}

so

t= 1hr

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