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3 years ago

19/32 reduced to its lowest form

Engineering

2 answers:

d1i1m1o1n [39]3 years ago

8 0

Answer:

19/32

Explanation:

19/32 is already in its lowest form

Iteru [2.4K]3 years ago

6 0

Answer:

0.59375 (rounded)

Explanation:

So 19/32 is already simplified since 19 is prime.

I assumed that you wanted to have it at it's simplest form?

So technically it's already at it's lowest form but I just gave you the decimal form.

Hopefully I was helpful!

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A cylinder contains 480 cm3 of loose dry sand which weighs 820 g. Under a static load of 200 kPa the volume is reduced 1%, and t

goblinko [34]

Answer:

Explanation:

a. void ratio is provided by the formula: $e = \frac{V_{p} }{V_{s} }$

where , $V_{p}$ = volume of voids

$V_{s}$ = volume of solid grains

for loose sand, the void space = $\frac{480}{480}$

= 1

b. void ratio after static load = 0.1/(480)/ (480)

= 0.1

c. void ratio after vibration = [480- ( 0.1 * 480) ]/ 480

= 0.9

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3 years ago

The lid on a pressure vessel is held down with 10 bolts that pass through the lid and a flange on the pressure vessel (similar t

Volgvan

Answer:

25120 N

Explanation:

The external load acting on each bolted joint

= P x A / N

= (8 x 10⁶) (31.4 x 10³ x 10⁻⁶) / 10

= 25120 N

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4 years ago

If the two 304-stainless-steel carriage bolts of the clamp each have a diameter of 10 mmmm, and they hold the cylinder snug with

S_A_V [24]

Answer: hello some data related to your question is missing attached below is the missing data

answer:

T2 = 265°C

Explanation:

First step : calculate sum of vertical forces

∑ y = 0

Fmg - 2(0.5 Fst ) = 0

∴Fmg = ( 12 * 10^6 ) ( 2 * π/4 (0.01)^2 )

= 1884.96 N

Also determine the Compatibility equation in order to determine the change in Temperature

ΔT = 250°C

therefore Temperature at which average normal stress becomes 12.0 MPa

ΔT = T2 - T1

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T2 = 250 + 15 = 265°C

attached below is the detailed solution

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3 years ago

Why less irrigation facilities in Telangana state give reasons

Paul [167]

Answer:

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3 0

3 years ago

Water flows in a pipeline. At a point in the line where the diameter is 7 in., the velocity is 12 fps and the pressure is 50 psi

PolarNik [594]

Answer:

a) P₂ = 3219.11 lbf / ft² , b) P₂ = 721.91 lbf / ft² , c) P₂ = 5707.31 lbf / ft²

Explanation:

For this exercise we can use the fluid mechanics equations, let's start with the continuity equation, index 1 is for the starting point and index 2 for the end point of the reduction

A₁ v₁ = A₂ v₂

v₂ = v₁ A₁ / A₂

The area of a circle is

A = π r² = π/4 d²

v₂ = v₁ (d₁ / d₂)²

Let's calculate

v₂ = 12 (7/3)²

v₂ = 65 feet / s

Now let's use Bernoulli's equation

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

P₁ - P₂ = ρ g (y₂ –y₁) + ½ ρ (v₂² - v₁²)

Case 1. The pipe is horizontal, so

y₁ = y₂

P₁ - P₂ = ½ ρ (v₂² –v₁²)

P₂ = P₁ - ½ ρ (v₂² –v₁²)

ρ = 62.43 lbf / ft³

P₁ = 50 psi (144 lbf/ ft² / psi) = 7200 lbf / ft²

P₂ = 7200 - ½ 62.43 / 32 (65² -12²)

P₂ = 7200 - 3980.89

P₂ = 3219.11 lbf / ft²

Case 2 vertical pipe with water flow up

y₂ –y₁ = 40 ft

P₁ - P₂ = ρ g (y₂ –y₁) + ½ rho (v₂² - v₁²)

7200 - P₂ = 62.43 (40) + ½ 62.43 / 32 (65 2 - 12 2) =

P₂ = 7200 - 2497.2 - 3980.89

P₂ = 721.91 lbf / ft²

Case 3. Vertical water pipe flows down

y₂ –y₁ = -40

P₂ = 7200 + 2497.2 - 3980.89

P₂ = 5707.31 lbf / ft²

3 0

3 years ago