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3 years ago

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Your heart pumps blood at a pressure of 100 mmHg and flow speed of 60 cm/s. At your brain, the blood enters capillaries with suc

h a large total cross-sectional area that the blood velocity is much smaller: about 0.7 mm/s. Ignore viscosity in the capillaries. How long could your neck be so that blood still reaches your brain in m

1 answer:

Savatey [412]3 years ago

7 0

Answer:

1.28 m

Explanation:

Generally, pressure of fluid is given by

$P=\rho g h$ where g is acceleration due to gravity, h is the height and $\rho$ is the density

Considering that the pressure for mercury is same as for blood only that the height and density of fluid are different then

$\rho_b g h_b= \rho_m g h_m$

Since g is constant, then

$\rho_b h_b= \rho_m h_m$

Making $h_b$ the subject of the formula then

$h_b=\frac {\rho_m h_m}{\rho_b}$

Where subscripts m and b denote mercury and blood respectively

Assuming density of blood is 1060 Kg/m3, density of mercury as 13600 Kg/m3 and substituting height of mercury for 0.1 m then

$h_b=\frac {13600*0.1}{1060}=1.283018868 m \approx 1.28 m$

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Suppose that a block of mass 2 kg is pulled to the right with a force of 10 N, and the friction force on the block is directed t

Law Incorporation [45]

Answer:

The block has an acceleration of $3 m/s^{2}$

Explanation:

By means of Newton's second law it can be determine the acceleration of the block.

$\sum F_{r} = ma$ (1)

Where $\sum F_{r}$ represents the net force, m is the mass and a is the acceleration.

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The forces present in x are $F = 10 N$ and $f = 4 N$ (the friction force):

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Notice that $f$ subtracts to $F$ since it is at the opposite direction.

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The forces present in y balance each other:

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Therefore:

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But $1 N = 1 Kg.m/s^{2}$ and writing (3) in terms of a it is get:

$a = \frac{6 Kg.m/s^{2}}{2 Kg}$

$a = 3 m/s^{2}$

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A simple harmonic oscillator consists of a block (m = 0.50 kg) attached to a spring (k = 128 N/m). The block is pulled a certain

Zinaida [17]

Answer:

0.5 m

14.00595

8 m/s, 0.0625 s

5.71314 m/s

Explanation:

k = Spring constant = 128 N/m

A = Amplitude

E = Energy in spring = 16 J

Energy in spring is given by

$E=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{2E}{k}}\\\Rightarrow A=\sqrt{\dfrac{2\times 16}{128}}\\\Rightarrow A=0.5\ m$

The amplitude is 0.5 m

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.5}{128}}\\\Rightarrow T=0.39269\ s$

Number of oscillations is given by

$N=\dfrac{5.5}{0.39269}\\\Rightarrow N=14.00595$

The number of oscillations is 14.00595

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$\dfrac{1}{2}mv^2=16\\\Rightarrow v=\sqrt{\dfrac{16\times 2}{0.5}}\\\Rightarrow v=8\ m/s$

The maximum speed is 8 m/s

For a distance of 0.5 m which is the amplitude

$Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{0.5}{8}\\\Rightarrow Time=0.0625\ s$

The time taken would be 0.0625 s

The maximum kinetic energy is equal to the mechanical energy

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At x = 0.35 m

$v=\sqrt{\dfrac{16-\dfrac{1}{2}kx^2}{\dfrac{1}{2}m}}\\\Rightarrow v=\sqrt{\dfrac{16-\dfrac{1}{2}128\times 0.35^2}{\dfrac{1}{2}0.5}}\\\Rightarrow v=5.71314\ m/s$

The speed of the block is 5.71314 m/s

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3 years ago

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