According to one government agency, bad housekeeping is estimated to cause what percentage

If you are trying to land the plane, which part(s) would you move? How would you move it/them?

KiRa [710]

Answer: hope it helps

Explanation:Moving air has a force that will lift kites and balloons up and down. Air is a mixture ... Here is a simple computer simulation that you can use to explore how wings make lift. ... All these dimensions together combine to control the flight of the plane. A pilot ... When the rudder is turned to one side, the airplane moves left or right.

7 0

4 years ago

For a bronze alloy, the stress at which plastic deformation begins is 275 MPa, and the modulus of elasticity is 115 GPa. (a) Wha

Anton [14]

Answer:

89375 N

Explanation:

Rearrange the formula for normal stress for F:

$\sigma=\frac{F}{A}$

$F=\sigma*A$

Convert given values to base units:

275 MPa = $275*10^{6}$ Pa

325 $mm^{2}$ = 0.000325 $m^{2}$

Substituting in given values:

F = $(275*10^{6})*(0.000325)=89375$ N

3 0

3 years ago

Consider three branch prediction schemes: predict not taken, predict taken, and dynamic prediction. Assume that they all have ze

SOVA2 [1]

Answer:

Check the explanation

Explanation:

1. When a branch is taken with 5% frequency, the branch prediction scheme is "predict taken" because the process has to begin fetching and then execute at the target address.

2. When a branch is taken with 95% frequency, the branch prediction scheme is "predict not taken" because the branch outcome is definitely known.

3. When a branch is taken with 70% frequency, the prediction scheme is "dynamic prediction" the branch prediction can dynamically change during the program execution.

3 0

3 years ago

**Please Help, ASAP**

Novay_Z [31]

Explanation:

We need to rearrange the following formula for the values given in parenthesis.

(1) x+xy = y, (x)

taking x common in LHS,

x(1+y)=y

$x=\dfrac{y}{1+y}$

(2) x+y = xy, (x)

Subrtacting both sides by xy.

x+y-xy = xy-xy

x+y-xy = 0

x-xy=-y

x(1-y)=-y

$x=\dfrac{-y}{1-y}$

(3) x = y+xy, (x)

Subrating both sides by xy

x-xy = y+xy-xy

x(1-y)=y

$x=\dfrac{y}{1-y}$

(4) E = (1/2)mv^2-(1/2)mu^2, (u)

Subtracting both sides by (1/2)mv^2

E-(1/2)mv^2 = (1/2)mv^2-(1/2)mu^2-(1/2)mv^2

E-(1/2)mv^2 =-(1/2)mu^2

So,

$2(E-\dfrac{1}{2}mv^2)=-mu^2\\\\u^2=\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)\\\\u=\sqrt{\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)}\\\\u=\sqrt{\dfrac{2}{m}(\dfrac{1}{2}mv^2-E)}$

(5) (x^2/a^2)-(y^2/b^2) = 1, (y)

$\dfrac{x^2}{a^2}-1=\dfrac{y^2}{b^2}\\\\y^2=b^2(\dfrac{x^2}{a^2}-1)\\\\y=b\sqrt{\dfrac{x^2}{a^2}-1}$

(6) ay^2 = x^3, (y)

$y^2=\dfrac{x^3}{a}\\\\y=\sqrt{\dfrac{x^3}{a}}$

Hence, this is the required solution.

3 0

3 years ago

1kg of air (R 287 J/kgK) fills a weighted piston-cylinder device at 50kPa and 100°C. The device is cooled until the temperature

Reika [66]

Answer:

the work done during this cooling is −28.7 kJ

Explanation:

Given data

mass (m) = 1 kg

r = 287 J/kg-K

pressure ( p) = 50 kPa

temperature (T) = 100°C = ( 100 +273 ) = 373 K

to find out

the work done during this cooling

Solution

we know the first law of thermodynamics

pv = mRT ....................1

here put value of p, m R and T and get volume v(a) when it initial stage in equation 1

50 v(a) = 1 × 0.287 × 373

v(a) = 107.051 / 50

v(a) = 2.1410 m³ .......................2

now we find out volume when temperature is 0°C

so put put value of p, m R and T and get volume v(b) when temperature is cooled in equation 1

50 v(b) = 1 × 0.287 × 273

v(a) = 78.351 / 50

v(a) = 1.5670 m³ .......................3

by equation 2 and 3 we find out work done to integrate the p with respect to v i.e.

work done = $\int\limits^a_b {p} \, dv$

integrate it and we get

work done = p ( v(b) - v(a) ) ................4

put the value p and v(a) and v(b) in equation 4 and we get

work done = p ( v(b) - v(a) )

work done = 50 ( 1.5670 - 2.1410 )

work done = 50 (−0.574)

work done = −28.7 kJ

here we can see work done is negative so its mean work done opposite in direction of inside air

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7 0

3 years ago