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Murljashka [212]
4 years ago
8

What is the scientific notation of 8,795,000

Mathematics
2 answers:
ryzh [129]4 years ago
8 0
8,795,000

8.795 x 1,000,000

(1,000,000 = 10*10*10*10*10*10 = 10^6 )

<u><em>8.795 x 10^6</em></u>
GuDViN [60]4 years ago
7 0
Answer: 8.795 * 10^6 = 8,795,000

------------------------------

8.795 * 10 = 87.95

8.795 * 10^2 = 879.5

8.795 * 10^3 = 8,795

8.795 * 10^4 = 87,950

8.795 * 10^5 = 879,500

8.795 * 10^6 = 8,795,000
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4 years ago
The volume of a cylinder is 500pi cm. The radius of the base of the cylinder is 5 cm. What is the height of the​ cylinder?
patriot [66]

Answer:

20cm

Step-by-step explanation:

Volume of cylinder = (pi) r^2 x height.

Rearrange:

500pi/5^2 pi = height

500pi/25pi = 20

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Hope this helped!

4 0
4 years ago
The rod is made of A-36 steel and has a diameter of 0.22 in . If the rod is 4 ft long when the springs are compressed 0.7 in . a
ziro4ka [17]

The force in the rod when the temperature is 150 °F is 718.72 pounds-force.

<h3>How to determine the resulting the resulting force due to mechanical and thermal deformation</h3>

Let suppose that rod experiments a <em>quasi-static</em> deformation and that both springs have a <em>linear</em> behavior, that is, force (F), in pounds-force, is directly proportional to deformation. Then, the elongation of the rod due to <em>temperature</em> increase creates a <em>spring</em> deformation additional to that associated with <em>mechanical</em> contact.

Given simmetry considerations, we derive an expression for the <em>spring</em> force (F), in pounds-force,  as a sum of mechanical and thermal effects by principle of superposition:

F = k\cdot (\Delta x + 0.5\cdot \Delta l)   (1)

Where:

  • k - Spring constant, in pounds-force per inch.
  • \Delta x - Spring deformation, in inches.
  • \Delta l - Rod elongation, in inches.

The <em>rod</em> elongation is described by the following <em>thermal</em> dilatation formula:

\Delta l = \alpha \cdot L_{o}\cdot (T_{f}-T_{o})   (2)

Where:

  • \alpha - Coefficient of linear expansion, in \frac{1}{^{\circ}F}.
  • L_{o} - Initial length of the rod, in inches.
  • T_{o} - Initial temperature, in degrees Fahrenheit.
  • T_{f} - Final temperature, in degrees Fahrenheit.

If we know k = 1000\,\frac{lb}{in}, \Delta x = 0.7\,in, \alpha = 6.5\times 10^{-6}\,\frac{1}{^{\circ}F}, L_{o} = 48\,in, T_{o} = 30\,^{\circ}F and T_{f} = 150\,^{\circ}F, then the force in the rod at final temperature is:

F = \left(1000\,\frac{lb}{in} \right)\cdot \left[0.7\,in + 0.5\cdot\left(6.5\times 10^{-6}\,\frac{1}{^{\circ}F} \right)\cdot (48\,in)\cdot (150\,^{\circ}F-30\,^{\circ}F)\right]

F = 718.72\,lbf

The force in the rod when the temperature is 150 °F is 718.72 pounds-force. \blacksquare

To learn more on deformations, we kindly invite to check this verified question: brainly.com/question/13774755

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Answer:

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