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zheka24 [161]
3 years ago
11

Please help, I don't get this

Physics
1 answer:
Trava [24]3 years ago
5 0
C the third one i think good luck
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The radius of the aorta is about 1 cm and the blood flowing through it has a speed of about 30 cm/s. Calculate the average speed
puteri [66]

Answer:

The average speed of the blood in the capillaries is 0.047 cm/s.

Explanation:

Given;

radius of the aorta, r₁ = 1 cm

speed of blood, v₁ = 30 cm/s

Area of the aorta, A₁ = πr₁² = π(1)² = 3.142 cm²

Area of the capillaries, A₂ = 2000 cm²

let the average speed of the blood in the capillaries = v₂

Apply continuity equation to determine the average speed of the blood in the capillaries.

A₁v₁ = A₂v₂

v₂ = (A₁v₁) / (A₂)

v₂ = (3.142 x 30) / (2000)

v₂ = 0.047 cm/s

Therefore, the average speed of the blood in the capillaries is 0.047 cm/s.

4 0
2 years ago
The S-strain Pneumococcus bacteria had a smooth surface, because ____________________.
drek231 [11]
The S strain Pneumococcus bacteria had a smooth surface because IT IS SURROUNDED BY A CARBOHYDRATE CAPSULE CALLED THE S STRAIN. The other form, the R strain has a rough surface and no capsule. It is only the S strain that exhibits virulence. 
4 0
3 years ago
A bird flies overhead from where you stand at an altitude of 270.0ĵ m and at a velocity horizontal to the ground of 14.0î m/is.
Varvara68 [4.7K]

Answer:

L = 8694 Kg.m²/s

Explanation:

r = 270 ĵ m

v = 14 î m/s

m = 2.3 kg

θ = 90º

L = ?

We can apply the equation

L = m*v*r*Sin θ

L = (2.3 kg)*(14 m/s)*(270 m)*Sin 90º = 8694 Kg.m²/s

8 0
3 years ago
What is meant by resistance and voltmeter?​
LUCKY_DIMON [66]

Answer:

<em>The internal resistance of an ideal ammeter will be zero since it should allow current to pass through it. Voltmeter measures the potential difference, it is connected in parallel. .</em>

Explanation:

<h3><em>I </em><em>hope</em><em> this</em><em> helps</em><em>!</em></h3>
7 0
2 years ago
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

x = 0.200 m

b)electric field at half way mean x =0.25

E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

E = 3.84*10^{-4} N/C

6 0
3 years ago
Read 2 more answers
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