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4 years ago

14

Which group of components is common to the circulatory systems of most living animals? A. arteries, veins, capillaries B. vessel

s, heart, circulating fluid C. aorta, ventricles, atria D. blood, heart, cavities

1 answer:

Artyom0805 [142]4 years ago

7 0

The circulatory system, composed of the heart, veins, arteries, capillaries, and blood, is a feature common in most vertebrates (animals with a distinct spine). In these animals, blood never leaves the confines of the heart and blood vessels, making theirs a closed circulatory system. In regard to your question, the best answer is B. Some animals have hearts that don’t have atria, while others don’t even have a distinct heart at all.

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MAVERICK [17]

10 would be the answer for number 3

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3 years ago

Two bulbs are connected in parallel across a source of EMF = 8.0V with a negligible internal resistance. One bulb has a resistan

Sonja [21]

<h2>Answer:</h2>

(a) 3.18Ω

(b) 3.18Ω

<h2>Explanation:</h2>

Let the two bulbs be A and B

Given;

$R_{A}$ = Resistance in bulb A = 3.0Ω

$R_{B}$ = Resistance in bulb B = 2.5Ω

Since the two bulbs are connected in parallel;

i. their effective resistance ( $R_{X}$ ) is given by

$\frac{1}{R_{X}}$ = $\frac{1}{R_{A} }$ + $\frac{1}{R_{B} }$ ---------------(i)

Substitute the values of $R_{A}$ and $R_{B}$ into equation (i)

=> $\frac{1}{R_{X}}$ = $\frac{1}{3.0}$ + $\frac{1}{2.5}$

Solve for $R_{X}$

$R_{X}$ = 1.36Ω

ii. voltage (potential difference), V, across them is the same;

Therefore we can get the total current (I) that will flow through them if the voltage to be supplied is 2.4V.

Use the Ohm's law;

V = I x R -----------------(ii)

Where;

V = voltage across them = 2.4V

I = total current flowing through them

R = their effective resistance = $R_{X}$ = 1.36Ω

Substitute these values into equation (ii) as follows;

2.4 = I x 1.36

I = 2.4 / 1.36

I = 1.76A

(a) Now get the value of R

Since the voltage across the two bulbs is 2.4V out of the 8.0V supplied by the source, then the remaining (8.0 - 2.4 = 5.6)V will pass across the resistor R.

Also, since the two bulbs make a series connection with the resistor R, the same total current (I = 1.76A) that flows through these bulbs will flow through the resistor R.

Therefore, to get the value of R, we use the relation

V = I x R ------------------------------(iii)

Where;

V = voltage across the resistor = 5.6V

I = current through the resistor = 1.76A

<em>Substitute these values into equation (iii)</em>

=> 5.6 = 1.76 x R

=> R = 5.6 / 1.76

=> R = 3.18Ω

Therefore, the value of R to be chosen in order to supply each bulb with a voltage of 2.4V is 3.18Ω

(b) The potential difference and voltage across refer to the same thing. Therefore, the value of R that would make the potential difference across each of the bulbs be 2.4V is the same as the one calculated in (a) which is 3.18Ω

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3 years ago

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3 years ago

What is dark energy?

expeople1 [14]

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4 0

3 years ago

A photon detector captures a photon with an energy of 4.29 ✕ 10−19 J. What is the wavelength, in nanometers, of the photon?

serious [3.7K]

Answer : The wavelength of photon is, $4.63\times 10^{2}nm$

Explanation : Given,

Energy of photon = $4.29\times 10^{-19}J$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency of photon

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength of photon = ?

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$4.29\times 10^{-19}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}$

$\lambda=4.63\times 10^{-7}m=4.63\times 10^{-7}\times 10^9nm=4.63\times 10^{2}nm$

Conversion used : $1nm=10^{-9}m$

Therefore, the wavelength of photon is, $4.63\times 10^{2}nm$

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3 years ago