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3 years ago

9

Water flows through a cylindrical pipe of varying cross-section. The velocity is 5.00 m/s at a point where the pipe diameter is

1.50 cm. At a point where the pipe diameter is 3.00 cm, the velocity is

1 answer:

NISA [10]3 years ago

5 0

Explanation:

We know that rate of flow through a cross section :

$v1 \times a1 = v2 \times a2$

5 m/s * 1.76cm^2 = v2 * 7.06cm^2

$v2 = 1.24 \: m {s}^{ - 1}$

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4 years ago

A balloon rises at the rate of 8 feet per second from a point on the ground 12 feet from an observer. To 2 decimal places in rad

Hitman42 [59]

Answer:

$\displaystyle \theta' =0.24\ rad/s$

Explanation:

<u>Rate Of Change</u>

Let some variable y depend on time t. we can express y as a function of t as

$y=f(t)$

The instant rate of change of y respect to t is the first derivative, i.e.

$y'=f'(t)$

The balloon, the ground and the observer form a right triangle (shown below) where the height of the balloon y, the horizontal distance x, and the angle of elevation are related with the trigonometric formula

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Solving for $\theta'$

$\displaystyle \theta' =\frac{y'}{xsec^2\theta}$

Let's compute the actual angle with the initial conditions y=9 feet, x=12 feet

$\displaystyle tan\theta =\frac{y}{x}$

$\displaystyle tan\theta =\frac{9}{12}=\frac{3}{4}$

Knowing that

$\sec^2\theta=1+tan^2\theta$

$\displaystyle \sec^2\theta=1+\left(\frac{3}{4}\right)^2$

$\displaystyle \sec^2\theta=\frac{25}{16}$

The balloon is rising at y'=8 feet/sec, thus we compute the change of the angle of elevation:

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$\boxed{\displaystyle \theta' =0.43\ rad/s}$

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So if you are trying to figure this question out you divide 24 by 5.1 to get your answer of 4.7 sec

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