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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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4 years ago

What is the temperature of a sample of gas when the average translational kinetic energy of a molecule in the sample is 8.37 × 1

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Answer:

404K

Explanation:

Data given, Kinetic Energy.K.E=8.37*10^-21J

Note: as the temperature of a is increase, the rate of random movement will increase, hence leading to more collision per unit time. Hence we can say that the relationship between the kinetic energy and the temperature is a direct variation.

This relationship can be expressed as

$K.E=\frac{3}{2}KT$

where K is a constant of value 1.38*10^-23

Hence if we substitute the values, we arrive at

$T=\frac{2/3(8.37*10^{-21})}{1.38*10^-23}\\ T=404K$

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3 years ago

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ZC. The forward force of the surfboard's acceleration is balanced by the backward force of the surfer's mass.

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4 years ago

1. An electron (Q=16x10^-20 C, m=1x10^-30 kg) moving at half a megameter per second up the page enters a region with a uniform m

jek_recluse [69]

Explanation:

It is given that,

Charge on electron, $q=16\times 10^{-20}\ C$

Mass of the electron, $m=9.1\times 10^{-31}\ kg$

Speed of the electron, $v=0.5\ Mm/s=0.5\times 10^6\ m/s$

Magnetic field, B = 1 T (directed out of the page)

Let F is the magnetic force acting on the electron. It is given by :

$F=qvB\ sin\theta$

Here, $\theta=90^{\circ}$

$F=qvB$

$F=16\times 10^{-20}\ C\times 0.5\times 10^6\ m/s\times 1\ T$

$F=8\times 10^{-14}\ N$

Using the right hand rule, the direction of magnetic force is upward to the plane of the paper. Also, the electron will follow the circular path. It is given by :

$r=\dfrac{mv}{qB}$

$r=\dfrac{9.1\times 10^{-31}\ kg\times 0.5\times 10^6\ m/s}{16\times 10^{-20}\ C\times 1\ T}$

$r=2.84\times 10^{-6}\ m$

Hence, this is the required solution.

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3 years ago