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pochemuha
3 years ago
11

A submarine deep underwater releases a bubble of air. The total pressure at this depth is P. The bubble has diameter, D, and abs

olute temperature, T. As the bubble rises, it expands. When it reaches a certain smaller depth, where the total pressure is 0.22P, it has absolute temperature 1.17. In terms of D, what is the diameter of the bubble at this depth? You may assume that no air leaves the bubble, and that the air behaves as an ideal gas.
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
7 0

Answer:

P V = n R T      ideal gas equation

P2 V2 / P1 V1 = T2 / T1    

V2 / V1 = T2 / T1 * P1 / P2 = T2 P1 / (T1 P2)

V2 / V1 = (1.17 T1) / T1 * (P1 / .22 P1)      assuming absolute temp as 1.17 P1

V2 / V1 = 1.17 / .22 = 5.32

V = 4/3 pi R^3 = 4/3 pi (D/2)^3 = 4/3 pi D^3 / 8 = pi D^3 / 6

V2 / V1 = D2^3 / D1^3

D2 = (V2 / V1 * D1^3)^1/3

D2 = 5.32^1/3 * D = 1.75 D      (D1 = D)

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A certain brand of hot-dog cooker works by applying a potential difference of 120 V across opposite ends of a hot dog and allowi
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Answer:

The time it will take to cook three hot dogs simultaneously is 2.5 minutes

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Here we have, the Energy of electric heating given by Joule heating that is;

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Since the energy required to cook one hotdog = 60.0 kJ we have

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The time it takes to cook three hot dogs simultaneously is

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5 0
3 years ago
A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w
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Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

\displaystyle P=\frac{W}{t}

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

W=F.x

The second Newton's law gives us the net force as

F=m.a

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

v_f=v_o+at

v_f=at

Solving for a

\displaystyle a=\frac{v_f}{t}={5.4}{1}

a=5.4\ m/s^2

The distance traveled in the interval is given by

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Since vo=0

\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}

x=2.7\ m

The force is given by

F=m.a

We don't know the value of m, so the force is

F=2.7m

Computing the work done by the sprinter

W=F.x=2.7m(5.4)

W=14.58m

The power is finally computed

\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}

P=14.58m

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}

a=8.8\ m/s^2

The distance is

\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}

x=3.8\ m

The net force is

F=m(8.8)=8.8m

The work done by the sprinter is now computed as

W=8.8m(3.8)=33.44m

At last, the output power is

\displaystyle P=\frac{33.44m}{0.5}=66.88m

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

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