Am I alive I really need to know?

Yfchggfcvkuhgvjbjhbsdfhjkibgihdfgb

Molodets [167]

Answer:

dysurygfyubiugyunhuhfnbgjdsbnm

Explanation: Because ngubduykrtvytiudhurthgvniuonssdf

8 0

3 years ago

Estimate horizontal and vertical tail sizes fo r a twin-turboprop regional transport with wing area of 40 m2, aspect ratio of 10

svetlana [45]

Answer: 0.2m sqr

Explanation:

A well behaved aircraft basically have a value of volume in horizontal and vertical area.

Volume in horizontal area (Vh) = 0.6

Volume in vertical area (Vv) = 0.05

Having known this, consider the relationship to find the vertical and horizontal tail sizes.

Vertical tail area (Sv)

Horizontal tail area (Sh)

Vh= (Sh × I) / S

Where,

I = moment

S= wing area

Sh= Horizontal tail area

Vh= Volume in horizontal area

0.6= Sh × 10/40

24= 10Sh

Sh= 24/10

Sh= 2.4 msqr

Horizontal tail area= 2.4m sqr

From the information above, we can calculate the vertical tail area.

Vertical tail area is calculated thus below:

Vv= (Sv× I) / S

Where

Vv= Volume in vertical area

Sv= Vertical tail area

I= Moment

S= Wing area

Therefore

Sv= (Vv × S) /I

Sv= (0.05×40)/10

Sv= 0.2msqr

In conclusion, the vertical tail size is 0.2msqr

7 0

3 years ago

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I am making composites of silicone rubber and copper particles; by mixing thermally conductive particles into the thermally insu

Gelneren [198K]

Answer:

The composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.

Explanation:

Let us assume that the total mass of composite is 100 lbm So as per the given conditions

15 lbm is copper and 85 lbm is rubber.
Density of rubber is 70 lbm/ft3
Specific gravity of Copper is 9

So As per the formula of specific gravity

$S_{cu}=\frac{\rho_{cu}}{\rho_w}$

here density of water is 62.4 lbm/ft3

Solving for Density of Copper gives

$S_{cu}=\frac{\rho_{cu}}{\rho_w}\\9=\frac{\rho_{cu}}{62.4}\\\rho_{cu}=9 \times 62.4\\\rho_{cu}=561.5 lbm/ft3$

For composition on volume basis, volume of individual components and composite are calculated as

$V_{cu}=\frac{m_{cu}}{\rho_{cu}}\\V_{cu}=\frac{15}{561.5}\\V_{cu}=0.0267 ft^3\\\\V_{r}=\frac{m_{r}}{\rho_{r}}\\V_{r}=\frac{85}{70}\\V_{r}=1.214 ft^3\\\\V_{c}=V_{r}+V_{cu}\\V_{c}=1.214+0.0267 \\V_{c}=1.2409 ft^3$

The composition is given as

$c_{cu}=\frac{V_{cu}}{V_{c}}\\c_{cu}=\frac{0.0267}{1.2409} \times 100 \%\\c_{cu}=2.154 \%\\\\c_{r}=\frac{V_{r}}{V_{c}}\\c_{r}=\frac{1.214}{1.2409} \times 100 \%\\c_{r}=97.83 \%$