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2 years ago

Do you know anything about Android graphics?

Engineering

1 answer:

Mashutka [201]2 years ago

7 0

Android provides a huge set of 2D-drawing APIs that allow you to create graphics.

Android has got visually appealing graphics and mind blowing animations.

The Android framework provides a rich set of powerful APIS for applying animation to UI elements and graphics as well as drawing custom 2D and 3D graphics.

<h3>Three Animation Systems Used In Android Applications:-</h3>

1. Property Animation

2. View Animation

3. Drawable Animation

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In order to avoid slipping in the shop, your footwear should __

madreJ [45]

Answer:

be like good

Explanation:

maybe rubber like so you don't slip

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3 years ago

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Water is contained in a rigid vessel of 5 m3 at a quality of 0.8 and a pressure of 1 MPa. If the pressure is reduced to 270.3 kP

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Answer:

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3 years ago

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3 years ago

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The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content

MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

$Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf$

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

$\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}$

Therefore, $V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}$

$V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy$

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since $1 yd^{3}= 27 ft^{3}$

The embankment requires water of 6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

$Unit_{weight}=\frac {17451720}{498594}=35.00186 lb$

1 gallon is approximately $8.35 yd^{3}$ hence

$\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}$

That's approximately 4.2 gallons

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3 years ago

A fuel cell vehicle draws 50 kW of power at 70 mph and is 40% efficient at rated power. You are asked to size the fuel cell syst

svetoff [14.1K]

Answer:

a) fuel cell weight and volume: 50 kg, 33.3 L

b) fuel tank weight and volume: 321 kg, 643 L

Explanation:

Delivery of 50 kW from a source with a power density of 1.5 kW/L requires a source that has a volume of ...

(50 kW)/(1.5 kW/L) = 33.3 L

The weight of the power source is 1 kW/kg, so will be ...

(50 kW)/(1 kW/kg) = 50 kg

400 miles at 70 mph will take (400/70) h ≈ 5.71429 h. In that time, the energy used by the vehicle power plant is ...

(50 kW)(5.71429 h)(3600 s/h) = 1028.57 MJ

Since the power plant is 40% efficient, it must be supplied with 2.5 times that amount of energy, or 2571.43 MJ.

The tank volume and mass will then be ...

volume = 2571.43 MJ/(4 MJ/L) = 642.9 L

mass = 2571.43 MJ/(8 MJ/L) = 321.4 kg

_____

<em>Comment on fuel tank volume</em>

It appears the fuel tank would need to be equivalent in size to a sphere about 1.1 m in diameter.

4 0

3 years ago