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pav-90 [236]
4 years ago
11

Explain the method to measure the external diameter of a sphere​

Physics
1 answer:
Serjik [45]4 years ago
8 0

Answer:

Sphere is a geometrical object in dimensional space  that surface of a circle and ball.

Explanation:

Sphere is that the circular objects in the two dimensional space (1) circle

(2) disk. Two dimensional space is a set of points and the distance of that point,The two points of Sphere that length and center.

Sphere can constructed as the named of surface form circle about  any diameter. circle is the special type of the revolution replacing the circle,

sphere is the distance r is the radius of the ball and circle  is the center of mathematical ball,as the center and the radius of the sphere is to respectively.

The ball and sphere has not be maintained mathematical references  as a solid references. A sphere of any radius is centered at the number of zero.

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Technician a says multiple discharge ignition system fires the spark plug during each of the engine's four cycle strokes. Techni
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4 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
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