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3 years ago

____________ is a medium-sized flat-topped hill with cliff-face sides that is taller than it is wide.

Physics

1 answer:

Luda [366]3 years ago

5 0

Answer:

mesa

Explanation:

A mesa is a flat-topped mountain or hill. It is a wide, flat, elevated landform with steep sides. ... Spanish explorers of the American southwest, where many mesas are found, used the word because the tops of mesas look like the tops of tables.

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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 $= 1.285 \times10^{-5} sec$ .

6 0

3 years ago

Help me Please!!!!!!!

Black_prince [1.1K]

The speed is equal to the area under the line up to the point where t .=15 s.
Do find the area of the triangle and that if the rectangle and add them together.
The area of the triangle is 25 and the rectangle is also 25 so the speed is 50 m/s

6 0

4 years ago

An electron in a TV picture tube is accelerated through a potential difference of 10 kV before it hits the screen. What is the k

laiz [17]

Answer:

10,000 eV

Explanation:

According to the law of conservation of energy, the kinetic energy gained by the electron is equal to its change in electric potential energy:

$K=\Delta U=q \Delta V$