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3 years ago

12

Help !

1 answer:

Black_prince [1.1K]3 years ago

7 0

First off, you need to know the weight of the projectile, lift and drag coefficients something like a high Reynolds number is preferred, then use the gravitational constant of 9.8 meters per second squared those would be a good start to get closer to your goal

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Which statements describe Rutherford model of the atom

il63 [147K]

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The nucleus is small and Very delicate. The electrons, for its turn it orbits around your nucleus. As the atom is the control/union of nucleus and electrons, and other particles, it is fair to say that an atom is a mostly empty space

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2 years ago

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What did hegel and kant have in common? they both were determined to disprove marxist theory. they both thought that only histor

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The two philosophers furthermore seem to share the same conception of the conditions of human freedom. For Hegel as well as Kant, a theory of morality and political right devoted to advancing the cause of freedom must require more than just the absence of obstacles preventing the satisfaction of our animal passions.

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1 year ago

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea

german

Answer:

Q1) Time taking by particle to travel the 40 km wrt. earth = $1.34\times10^{-6}$ sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = $1.285\times10^{-5}$ sec.

Explanation:

Given :

Speed of particle wrt. earth $v=0.99537c$

Distance between where particle is created and earth surface = 40 km

we know that,

⇒ $v = \frac{x}{t}$

Where $x = 40\times10^{3} m$ , $v = 0.99537c$ , we know speed of light $c = 3 \times10^{8}$

∴ $t = \frac{x}{v}$

$= \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }$

$t = 1.34\times10^{-6} sec$

∴ Thus, time taking by particle to travel the 40 km wrt. earth $t = 1.34\times10^{-6} sec$

According to the lorentz transformation,

⇒ $l = l_{o} \sqrt{1-\frac{v^2}{c^2} }$

Where $l =$ improper length, $l_{o} =$ proper length (distance measured wrt. rest frame) = 40 km

$l = 40 \sqrt{1-\frac{v^2}{c^2 }$

$l = 40 \times 0.096$

$l = 3.84$ km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

$\Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }$

Where $\Delta t =$ improper time (wrt. earth frame time) $=1.34\times10^{-6} sec$ , $\Delta t _{o} =$ proper time (wrt. particle frame).

$1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}$

$\Delta t_{o} = 1.285 \times10^{-5} sec$

Thus, the time taking by particle to travel from where it is created to the surface of the earth $= 1.285 \times10^{-5} sec$ .

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2 years ago

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AnnyKZ [126]

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2 years ago

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3 years ago