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2 years ago

What type of siege engines were used by Saladin to capture Jerusalem in 1187?

Engineering

1 answer:

mariarad [96]2 years ago

8 0

Answer:

LOTS

Explanation:

Catapults, Towers, and Trebuchets were all used by Saladin to capture Jerusalem in 1187

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What is the weight density of a 2.24 in diameter titanium sphere that weights 0.82 lb?

nasty-shy [4]

Answer:

$0.14\ lb/in^{3}$

Explanation:

Density is defined as mass ler unit volume, expressed as

$\rho=\frac {m}{v}$

Where m is mass, $\rho$ is density and v is the volume. For a sphere, volume is given as

$v=\frac {4\pi r^{3}}{3}$

Replacing this into the formula of density then

$\rho=\frac {m}{\frac {4\pi r^{3}}{3}}$

Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then

$\rho=\frac {0.82}{\frac {4\pi 1.12^{3}}{3}}=0.13932044952427\approx 0.14 lb/in^{3}$

4 0

3 years ago

A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob

scoray [572]

Solution :

Given :

The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

$= 71.82 \ kN/m^2$

$\sim 72 \ kN/m^2$

Now,

$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$

$N_1 =$ corrected N - value of overburden

$\bar{\sigma}=$ effective stress at level of test

0 - 6 inch, $N_1=4 \left(\frac{350}{72+70} \right)$

= 9.86

6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

$N_{avg}=\frac{9.86+14.8+14.8}{3}$

= 13.14

= 13

8 0

2 years ago

A gearbox is needed to provide an exact 30:1 increase in speed, while minimizing the

alekssr [168]

Answer:

answer

Explanation:

4 0

2 years ago

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sukhopar [10]

Answer:

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Explanation:

6 0

2 years ago

A circular column is fixed at the base and not supported at the top. If the column needs to be 15ft and hold 10kips, what is the

muminat

Answer:

The required size of column is length = 15 ft and diameter = 4.04 inches

Explanation:

Given;

Length of the column, L = 15 ft

Applied load, P = 10 kips = 10 × 10³ Psi

End condition as fixed at the base and free at the top

thus,

Effective length of the column, $\L_e$ = 2L = 30 ft = 360 inches

now, for aluminium

Elastic modulus, E = 1.0 × 10⁷ Psi

Now, from the Euler's critical load, we have

$P =\frac{\pi^2EI}{L_e^2}$

where, I is the moment of inertia

on substituting the respective values, we get

$10\times10^3 =\frac{\pi^2\times1.0\times10^7\times I}{360^2}$

I = 13.13 in⁴

also for circular cross-section

I = $\frac{\pi}{64}\times d^4$

thus,

13.13 = $\frac{\pi}{64}\times d^4$

d = 4.04 inches

The required size of column is length = 15 ft and diameter = 4.04 inches

3 0

2 years ago