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3 years ago

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that delivers oxygen to your body and In the video your blood is compared to a picks up CO2 to be released out when you breath.

PLEASE I NEED A ANSWER

1 answer:

sweet-ann [11.9K]3 years ago

4 0

I’m a bit confused. What’s the question?

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A well hole having a diameter of 5 cm is to be cut into the earth to a depth of 75 m. Determine the total work (in joules) requi

guapka [62]

Answer:

total work is 99.138 kJ

Explanation:

given data

diameter = 5 cm

depth = 75 m

density = 1830 kg/m³

to find out

the total work

solution

we know mass of volume is

volume = $\frac{\pi}{4} d^2 dx$

volume = $\frac{\pi}{4} d^2 1830 dx$

so

work required to rise the mass to the height of x m

dw = $\frac{\pi}{4} d^2 1830$ gx dx

so total work is integrate it with 0 to 75

w = $\int\limits^{75}_{0} {\frac{\pi}{4} d^2 1830 gx dx}$

w = $\frac{\pi}{4}$ × 0.05² × 1830× 9.81× $(\frac{x^2}{2})^{75}_0$

w = 99138.53 J

so total work is 99.138 kJ

6 0

2 years ago

A scientist adds different amounts of salt to 5 bottles of water. She then measures how long it takes for the water to boil. Wha

postnew [5]

Answer:

the responding variable is the water boiling

Explanation:

a responding variable is the same thing as a dependent variable and an independent variable you change the independent variable is the amount of salt, the control group is how long water takes to boil without adding salt, and a constant is the same amount of water

8 0

2 years ago

A water tank is in the shape of an inverted cone with depth 10 meters, and top radius 8 meters. Water is flowing into the tank a

Elis [28]

Answer:

The value of leaking rate in the question is repeated. By searching on the web I could find the correct value wich is 0.002h^2 m^3 /min.

The depth of the water has to be equal to 7.07 m in order to have a stationary volume.

Explanation:

In order to have a stationary water level the flow of water that comes into the tank (0.1 m^3/min) must be equal to the flow of water that goes out of the tank (0.002*h^2 m^3/min), therefore:

0.002*h^2 = 0.1

h^2 = 0.1/0.002

h^2 = 50

h = sqrt(50) = 7.07 m

7 0

2 years ago

Two organ pipes, both open at one end but closed at the other, are each 1.14 m long. One is now lengthened by 2 cm. Find the fre

zloy xaker [14]

Answer: 1.3 Hz

Explanation:

4 0

3 years ago

A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o

blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

$P=\rho \times g \times h$

Here

P is the pressure which is given as 68 kPa.
ρ is the density of the oil whose SG is 0.92. It is calculated as

$\rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\$

g is the gravitational constant whose value is 9.8 m/s^2
h is the height which is to be calculated

$P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m$

So the height of column is 7.54m

a-3

By the relation of volume and density

$M=\rho \times V$

Here

ρ is the density of the oil which is 920 kg/m^3
V is the volume of cylinder with diameter 16m calculated as follows

$V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3$

Mass is given as

$M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg$

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

$\Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\$

Now corrected pressure is as

$P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa$

Finding the value of height for this corrected pressure as

$P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m$

The original height of column is 5.98m

4 0

3 years ago