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Amanda [17]
3 years ago
12

One Pro for supplements is the cost. True Or False

Physics
1 answer:
Nat2105 [25]3 years ago
8 0

Answer:

False

Explanation:

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Human blood consists of blood groups A, B, and O. However, there is one more blood group, AB, which contains alleles A and B in
Arturiano [62]
The answer is Codominance, AB blood cells inherit both A and B blood types so they are codominance
3 0
3 years ago
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How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t
Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

Qo=1.6 \times 10^{2} \mathrm{~mL}

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}

$$

Where dP is pressure difference r is the radius

\mu is the viscosity of water

L is the length of the pipe

Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}

In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

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5 0
2 years ago
A 6.8 kg bowling ball and 7.4 kg bowling ball rest on a rack 0.74 m apart. What is the force of gravity pulling each ball toward
zimovet [89]

The gravitational force between the two balls is 6.13\cdot 10^{-9} N

Explanation:

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where :

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

For the balls in this problem,  we have

m_1 = 6.8 kg

m_2 = 7.4 kg

r = 0.74 m

Substituting into the equation, we find the gravitational force between the two balls:

F=(6.67\cdot 10^{-11})\frac{(6.8)(7.4)}{(0.74)^2}=6.13\cdot 10^{-9}N

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4 0
3 years ago
How does changes in distance affect the gravitational pull between two objects? Describe and give one example.
maxonik [38]
The formula is

F_grav = G * m1 * m2 / r^2

G m1 and m2 are going to stay the same once chosen no matter what the distance is. The only thing that will change is the distance.

As the distance increases, the Gravitational Force will decrease. It will decrease by quite a bit.

As the distance decreases, the gravitational force will Increase.

The relationship is inverse. The moon travelling around the earth is one example. The earth travelling around the sun is another.
8 0
3 years ago
The force of air particles over an area is A.) temperature B.) pressure C.) Volume D.) Kelvins
Westkost [7]

Answer:The anwser is B i Promise ok

Explanation:

3 0
3 years ago
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