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Elena-2011 [213]

3 years ago

11

As blood passes through the capillary bed in an organ, the capillaries join to form venules (small veins). If the blood speed in

creases by a factor of 4.00 and the total cross-sectional area of the venules is 10.0cm2, what is the total cross-sectional area of the capillaries feeding these venules?

1 answer:

Semenov [28]3 years ago

4 0

Answer:

$A_2 = 2.5\ cm^2$

Explanation:

given,

velocity factor = 4

Cross-sectional area of venules(A₁) = 10 cm²

cross sectional area of capillaries(A_2) = ?

continuity equation = Q = AV

now,

$\dfrac{A_1}{A_2} = \dfrac{V_2}{V_1}$

$\dfrac{10}{A_2} =4$

$\dfrac{10}{4} =A_2$

$A_2 = 2.5\ cm^2$

hence, the area of capillaries is equal to 2.5 cm₂

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The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.4 V is across it at a frequ

zzz [600]

Answer:

The value of the inductance is 1.364 mH.

Explanation:

Given;

amplitude current, I₀ = 200 mA = 0.2 A

amplitude voltage, V₀ = 2.4 V

frequency of the wave, f = 1400 Hz

The inductive reactance is calculated;

$X_l = \frac{V_o}{I_o} \\\\X_l = \frac{2.4}{0.2} \\\\X_l =12 \ ohms$

The inductive reactance is calculated as;

$X_l = \omega L\\\\X_l = 2\pi fL\\\\L = \frac{X_l}{2 \pi f}$

where;

L is the inductance

$L = \frac{12}{2 \pi \times \ 1400} \\\\L = 1.364 \times \ 10^{-3} \ H\\\\L = 1.364 \ mH$

Therefore, the value of the inductance is 1.364 mH.

7 0

3 years ago

A ball is thrown straight up with enough speed so that it is in the air for several seconds. Assume the positive direction is up

Andrew [12]

Answer:

a) v= 0 m/s b) v= 6.86 m/s

Explanation:

a) When the ball reaches to its highest point, under the influence of gravity, before starting to fall down, it momentarily comes to an stop (this is needed prior to change direction in any movement), so, applying the definition of acceleration, and replacing the acceleration a by g, we have:

vf = v₀ - g*t (1)

The minus sign means that the acceleration due to gravity is always downward, so if we assume that the positive direction is upwards it must be negative.

At the highest point, vf= 0.

b) Prior to solve this point, we need to know which is the time when the ball reaches to its highest point.

As we know vf=0, we can solve (1) for t, as follows:

th = v₀/g

Now, for a time that is 0.7 s before this time, applying the acceleration definition and solving for v again, we have:

v = v₀ -(g *(th-0.7 s)), but th= v₀/g, so we get:

v= v₀ -g((v₀/g)-0.7 s) = v₀ - v₀ + g*0.7 s

⇒ v=g*0.7 s = 9.8 m/s²*0.7 s

⇒ v = 6.86 m/s

6 0

3 years ago

marishachu [46]

Answer:

The box will experience an acceleration.

Explanation:

Here, 2 N and 3 N forces are acting opposite to each other. In this case, the net force experience by the box would be (3-2)N = 1 N towards right. Since acceleration is directly proportional to the net force, therefore the box will experience an acceleration.

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3 years ago

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g

Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

$k=\sqrt{\frac{I}{M} }$

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration $k=\sqrt{\frac{I}{M} }$

So, moment of rotational inertia (I) of a cylinder about it axis = $\frac{MR^2}{2}$

$k=\sqrt{\frac{\frac{MR^2}{2}}{M} }$

$k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }$

$k=\sqrt{{\frac{R^2}{2}}$

$k={\frac{R}{\sqrt{2}}$

$k={\frac{1.20m}{\sqrt{2}}$

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = $\frac{2}{3}MR^2$

$k = \sqrt{\frac{\frac{2}{3}MR^2}{M} }$

$k = \sqrt{\frac{2}{3} R^2}$

$k = \sqrt{\frac{2}{3} }*R$

$k = \sqrt{\frac{2}{3}} *1.20$

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of radius

(I) = $\frac{2}{5}MR^2$

$k = \sqrt{\frac{\frac{2}{5}MR^2}{M} }$

$k = \sqrt{\frac{2}{5} R^2}$

$k = \sqrt{\frac{2}{5} }*R$

$k = \sqrt{\frac{2}{5}} *1.20$

k = 0.7560

k ≅ 0.76 m

6 0

3 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

MArishka [77]

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

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3 years ago