If i eat myself will I get twice as big or disappear completely?

Consider the following grooves, each of width W, that have been machined from a solid block of material. (a) For each case obtai

kogti [31]

Answer:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

8 0

3 years ago

Which is equal to a temperature of 50°F? 18°C 46°C 10°C 32°C

Ludmilka [50]

Answer:

10°C degrees po ang sagot

8 0

2 years ago

Read 2 more answers

An electrical heater is a form of sensible heating process, and heats 0.1m/s of air from 15°C and 80% RH to 50°C? The barometric

lawyer [7]

Answer:

The heater load =35 KJ/kg

Explanation:

Given that

At initial condition

Temperature= 15°C

RH=80%

At final condition

Temperature= 50°C

We know that in sensible heating process humidity ratio remain constant.

Now from chart

At temperature= 15°C and RH=80%

$h_1=38 \frac{KJ}{kg},v=0.8 \frac{m^3}{kg}$

At temperature= 50°C

$h_2=73 \frac{KJ}{kg}$

$So\ the\ heater\ load =h_2-h_1$

The heater load = 73 - 38 KJ/kg

The heater load =35 KJ/kg

3 0

2 years ago

To find the quotient of 8 divided by 1/3, multiply 8 by?

iVinArrow [24]

Answer: 8.33333333 or 6.1989778

Explanation:

8 0

2 years ago

A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur

anyanavicka [17]

Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603 225

.203 368

.162 442

answer : 191 mPa

Explanation:

Determine the stress present at the time of fracture for the original plate

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603 bending stress = 225

at 0.203 bending stress = 368

at 0.162 bending stress = 442

applying equation 1 determine the value of n for several combinations

( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) = ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x = 0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

= 225 * 0.84747

= 191 mPa 

3 0

2 years ago