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3 years ago

What is the danger to spacecraft and astronauts from micrometeoroids?

Physics

1 answer:

Sloan [31]3 years ago

7 0

Answer:

Suppose the micrometeoroid weighed 1 g = .001 kg

Suppose also the spacecraft were moving at 18,000 mph (1.5 hrs per rev)

Usually, the smaller particle would be moving but for simplicity suppose that it were stationary wrt the ground

v = 18000 miles / hr * 1500 m/mile / 3600 sec/hr = 7500 m/s

KE = 1/2 * .001 kg * (7500 m)^2 = 28,125 Joules

One can see that 28000 Joules could be damaging amount of energy

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Find the weight of an object of mass 5 kg

Elis [28]

Answer:

weight on earth is mg

which is 5*9.8

49 Newton

weight on moon is 1/6 th of weight on earth

1/6*49

8.166 Newton..

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3 years ago

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EXPLAIN HOW ENERGY IS TRANSMITTED THROUGH A MEDIUM

jek_recluse [69]

I say it helped then because TrueType had room

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4 years ago

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The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of

natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

$\mu$ = Coefficient of friction = 0.4

Energy stored in spring is given by

$U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J$

As the energy in the system is conserved we have

$\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s$

The speed of the 8 kg block just before collision is 3.258 m/s

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3 years ago

There are some points on a standing

bija089 [108]

Answer:

Nodes.

Explanation:

Nodes are a point that are on a standing wave that never move.

7 0

4 years ago

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A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is

harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

Where,

$d_g =$ Depth of glass

$n_w =$ Refraction index of water

$n_g =$ Refraction index of glass

$n_{air} =$ Refraction index of air

$d_w =$ Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

$d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})$

$d'w = 4.041cm$

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0

3 years ago