6

Blood contains positive and negative ions and therefore is aconductor. A blood vessel, therefore, can be viewed as anelectrical

CaHeK987 [17]

Answer:

<h2>Magnetic field required for the given induced EMF is 1.41 T</h2>

Explanation:

Potential difference across the blood vessel is given as

$E = vBd$

here we know that the speed is given as

$v = 14.8 cm/s$

$d = 4.80 mm$

$E = 1 mV$

now we have

$1 \times 10^{-3} = (14.8 \times 10^{-2})B(4.80 \times 10^{-3})$

$B = 1.41 T$

Now volume flow rate of the blood is given as

$Q = Av$

$Q = \frac{\pi d^2v}{4}$

from above equation we have

$v = \frac{E}{Bd}$

Now we have

$Q = \frac{\pi d^2\frac{E}{Bd}}{4}$

$Q = \frac{\pi E d}{4B}$

5 0

3 years ago

A ball is launched from ground level and hits the ground again after an elapsed time of 4 seconds and after traveling a horizont

jeka94

Answer:

1)a. It is constant the whole time the ball is in free-fall.

2)b. = 14 m/s

3) e. = 19.6 m/s

Explanation:

1) given that the only force acting on the ball is gravity, gravity acts along the vertical axis. Since no other force acts on the ball then the horizontal velocity will remain constant all through the flight since there is no horizontal force acting on the ball.

2) speed = distance/time

horizontal distance = 56m

Time = 4 seconds

Speed = 56m/4s = 14m/s

3) acceleration due to gravity g = 9.8m/s^2

Initial vertical velocity = u

Final vertical velocity = v = -u

Using the law of motion;

v = u + at

a = acceleration = -g = -9.8m/s^2

t = time of flight = 4

Substituting the values;

-u = u - 4(9.8)

-2u = -4(9.8)

u = -4(9.8)/-2

u = 2(9.8) = 19.6 m/s

Initial vertical velocity = u = 19.6 m/s

3 0

3 years ago

State two factors that affect the rate of diffusion of a substance

Ronch [10]

Explanation:

Two factors that affect the rater of diffusion of a substance are:

Diffusion of substance plays an important role on cellular transport in plants.
Diffusion is the passive movement of substance from a region of higher concentration to a region of lower concentration.

4 0

3 years ago

A 5 kg block is resting on a ramp inclined at 35 degrees above the horizontal. What is the magnitude of the normal force acting

Mamont248 [21]

I'm pretty sure the answer is b 28n hope helps :)

4 0

3 years ago

When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the

MaRussiya [10]

Answer:

(a) $\alpha=-111.26rad/s$

(b) $s=4450.6in$

(c) $8.66in$

Explanation:

First change the units of the velocity, using these equivalents $1rev=2\pi rad$ and $1 min =60s$

$4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s$

The angular acceleration $\alpha$ the time rate of change of the angular speed $\omega$ according to:

$\alpha=\frac{\Delta \omega}{\Delta t}$

$\Delta \omega=\omega_i-\omega_f$

Where $\omega_i$ is the original velocity, in the case the velocity before starting the deceleration, and $\omega_f$ is the final velocity, equal to zero because it has stopped.

$\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s$

b) To find the distance traveled in radians use the formula:

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

$\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad$

To change this result to inches, solve the angular displacement $\theta$ for the distance traveled $s$ ( $r$ is the radius).

$\theta=\frac{s}{r} \\s=\theta r$

$s=890.12(5)=4450.6in$

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

$\frac{890.12}{2\pi}=141.6667$

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle $\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120$ is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):

$c^2=a^2+b^2-2abcos(\gamma)$

$c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in$

4 0

4 years ago