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3 years ago

15

If cast iron is such a hard metal why dont the military use cast iron to armor their tanks? Is it cuz cast iron is such a heavy

metal that itll make the tank slow? Or is there not enough of cast iron on earth?

1 answer:

Stels [109]3 years ago

6 0

Weight is a really important factor in military applications. tanks need light durable and these days intelligent armour systems. most armoured personel carriers only have aluminium walls. Tanks these days have reactive armour and laser systems to neutralise incomming threats, rather than just relying on heavy armour.

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A few alpha particles bounce off a thin sheet of gold foil. What did scientists

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I world say A is the answer 80% sure

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2 years ago

Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh

MariettaO [177]

Answer:

The separation between the two lowest levels = $1.24 * 10^{-39}J$

The values of n where the energy of molecule reaches 1/2 kT at 300K = $2.2 * 10^{9}$

The separation at this level = 1.8 * $10^{-30}$ J

Explanation:

Knowing the formula

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$

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E2 - E1 = $\frac{3h^{2} }{8mL^{2} }$

E2 - E1 = $\frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2} } = 1.24 * 10^{-39}J$

Now we should find n where the energy of molecule reaches 1/2 kT

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$ = $\frac{1}{2}kT$

$\frac{h^{2} }{8 mL^{2} }$ = $4.13 * 10^{-14}J$

$n^{2} * (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K$

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3 years ago

A block of ice is sliding down a ramp of slope 40 to the horizontal. What is the rate of acceleration of the block? Assume the f

uysha [10]

The given problem can be exemplified in the following diagram:

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$mg\sin 40=ma$

We may cancel out the mass:

$g\sin 40=a$

Using the gravity constant as 9.8 meters per square second:

$9.8\frac{m}{s^2}\sin 40=a$

Solving the operations:

$6.3\frac{m}{s^2}=a$

Therefore, the acceleration is 6.3 meters per square second.

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1 year ago

An electron initially has a speed 16 km/s along the x-direction and enters an electric field of strength 27 mV/m that points in

weqwewe [10]

Answer:

a) $t=1.4\times 10^{-5}\ s$

b)S= 46.4 cm

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d= 22.5 cm

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b)

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Here initial velocity u= 0 m/s

$S= \dfrac{1}{2}\times 4.74\times 10^{9}\times (1.4\times 10^{-5})^2\ m$

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S is the deflection of electron.

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