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3 years ago

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At its widest point, the diameter of a bottlenose dolphin is 0.50 m. Bottlenose dolphins are particularly sleek, having a drag c

oefficient of only about 0.090. a. What is the drag force acting on such a dolphin swimming at 7.5 m/s? b. Using the dolphin’s diameter as its characteristic length, what is the Reynolds number as it swims at this speed in 20° C water?

1 answer:

fiasKO [112]3 years ago

8 0

Answer:

497.00977 N

3742514.97005

Explanation:

$\rho$ = Density of water = 1000 kg/m³

C = Drag coefficient = 0.09

v = Velocity of dolphin = 7.5 m/s

r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m

A = Area

Drag force

$F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N$

The drag force on the dolphin's nose is 497.00977 N

at 20°C

$\mu$ = Dynamic viscosity = $1.002\times 10^{-3}\ Pas$

Reynold's Number

$Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005$

The Reynolds number is 3742514.97005

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Answer:

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Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .

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Average EMF, which is the same as the average rate of change in flux:

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3 years ago

si un disco fue lanzado con una fuerza de 1000 N, y el disc recorrió una distancia e o.6 m ¿Qué trabajo efectuó el lanzador ?

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Answer:

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Explanation:

Dados los siguientes datos;

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2 years ago

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Answer:

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3 years ago