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3 years ago

Why is the definition for second based on the Cesium 133 atom?

Engineering

1 answer:

jarptica [38.1K]3 years ago

7 0

Answer:

The rate of vibrations of the atoms is consistent

Explanation:

A second is defined as 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom.

The definition applies when the atom is in isolation at a temperature of absolute zero. Adjustments must be made for the conditions that can be maintained in practice.

Such radiation is determined by the laws of quantum mechanics, so is extremely consistent.

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Explain the differences between planned and predictive maintenance.

sveticcg [70]

Answer:

Planned maintenance refers to any scheduled activity carried out to check a machine is working ok and diagnose procedures to fix it if need it. On the other hand, predictive mainteance is all the techniques which help to define if a machine requires or not maintenance activities so far.

Explanation:

Planned maintenance is based on preventive routines to ensure a machine is working in acceptable conditions and at the same time prevent them to change to risky values performing acticities like parts replacement, cleaning, etc. The key of this maintenance is schedule, that is to say, is a maintenance that has to be carried out constantly each certain time. Predictive maintenance is different because it is used to define if a machie needs any kind of inspection or if, on the contrary, the machine can continue operating without any intervention. The good point about predictive maintenance is the capability of telling when a maintenance is required and when is no necessarily required which is ideal to save costs.

7 0

3 years ago

1. A thin-walled cylindrical pressure vessel is capped at the end and is subjected to an internal pressure (p). The inside diame

Vesna [10]

I DONT KNOW OKAY UGHHH

6 0

3 years ago

Which of the eight diagnostic steps for locating an engine performance problem is performed first?

Kay [80]

Answer:

D. Perform a thorough visual inspection.

4 0

2 years ago

g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,

irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0

3 years ago

Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given

Akimi4 [234]

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = $\frac{1}{6}*334.8 = 55.8 ft^3$

Volume of sand = $\frac{2}{6}*334.8 = 111.6 ft^3$

Volume of gravel = $\frac{3}{6}*334.8 = 167.4 ft^3$

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

= 1/2 * 10968 = 5484 lb

4 0

3 years ago