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Marrrta [24]
3 years ago
14

There is a limit to how long your neck can be. If your neck were too long, no blood would reach your brain! What is the maximum

height a person's brain could be above his heart, given the noted pressure and assuming that there are no valves or supplementary pumping mechanisms in the neck? The density of blood is 1060 kg/m3 .
Physics
1 answer:
spayn [35]3 years ago
5 0

Answer:

The maximum height a person's brain could be above his heart is: 1.28 meter.

Explanation:

We need to know what is the normal blood pressure ours hearts so there is a rate: 120/80 (mmHg) and the average will be: 100 (mmHg) and using the Pascal law that relate pressure, density, gravity and height like:P_{2} = pgh_{1} - pgh_{2} + P_{1}, where P is pressure, p is density, g is the gravity acceleration and h is the height. Now we can find the height and delta of pressure will be: P2-P1 = 100 (mmHg), knowing that 1(mmHg) is equal to 133 Pa, we can do the convertion to 13332.2 (Pa), now because the units of Pascal are Newton/(meter^2). Then we solve the formula to get the height: \frac{P2-P1}{pg} =h so we get:\frac{13332.2}{(1060*9,81)}=Height=1.28(meters).

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5 0
3 years ago
330 g of water at 55°c are poured into an 855 g aluminum container with an initial temperature of 10°
Olenka [21]
The final temperature of the system is 32.5°
we know,  H = mcT 
where, H = Heat content of the body 
m = Mass,
c = Specific heat
T = Change in temperature
According to to the Principle of Calorimetry 

The net heat remains constant i.e. 
⇒ the heat given by water = heat accepted by the aluminum container.
⇒ 330 x 1 x (45 - T) = 855 x \frac{900}{4200} x (T - 10) 

⇒ 14,850 - 330T = 183.21T  - 1832 

⇒ - 513.21 T = - 16682

or T = 32.5°
3 0
3 years ago
A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the
podryga [215]

Answer:

6.9\times 10^{28}m^{-3}

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=rr=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m

1mm=10^{-3} m

Current=I=8 A

Drift velocity=v_d=5.4\times 10^{-5} m/s

We have to find the density of free electrons in the metal

We know that

Density of electron=n=\frac{I}{v_deA}

Using the formula

Density of free electrons=\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}

By using Area of wire=\pi r^2

\pi=3.14\\e=1.6\times 10^{-19} C

Density of free electrons=6.9\times 10^{28}m^{-3}

3 0
3 years ago
Read 2 more answers
The number of significant figures in 0.060900 is
Lady bird [3.3K]

Answer:

There are <u>5</u> significant figures.

Explanation:

You must start conting your sig figs until you continue to hit zeros at the end. Those zeroes at the end are disregarded. So 0.0609 is where you get your <em>sig figs</em> from.

8 0
3 years ago
Identical point charges (+50 x 10 power -6C) are placed at the corners of a square with sides of 2.0-m length. How much external
Gnesinka [82]

Answer:

636.4 J

Explanation:

The potential energy between one of the charges at the corner of the square and the fifth identical charge is U = kq²/r where q = charge = +50 × 10⁻⁶ C  and r = distance from center of square. = √2 m (since the midpoint of the sides = 1 m, so the distance from the charge at the corner to the center is thus √(1² + 1²) = √2)

Since we have four charges, the additional potential energy to move the charge to the centre of the square is U' = 4U = 4kq²/r

U' = 4kq²/r

= 4 × 9 × 10⁹ Nm²/C² (+50 × 10⁻⁶ C)²/√2 m

= 900 Nm²/√2 m

= 636.4 J

8 0
4 years ago
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