Each step in any energy conversion process will ____

Fr-ee p-o-i-n-t-s is biden bad yes or no and tell me why

dalvyx [7]

Answer:

Wohhh thanks for you free answer. Well i will be appreciating it.

5 0

3 years ago

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Tyson Gay's best time to run 100.0 meters was 9.69 seconds. What was his average speed during this run, in miles per hour? (3.28

FinnZ [79.3K]

Answer:

23.086 mile/h

Explanation:

Given,

Distance Tyson Gay run = 100 m

time of run, t = 9.69 s

average speed of the in mph = ?

Speed of the Gay = $\dfrac{distance}{time}$

$v = \dfrac{100}{9.69}$

v = 10.32 m/s

1 m = 3.281 ft

10.32 m = 33.86 ft

1 mile = 5280 ft

1 ft = 1.8939 x 10⁻⁴ mile

33.86 ft/s = 6.413 x 10⁻³ miles/s

Speed of Tyson in mile/hr = 6.413 x 10⁻³ x 3600

= 23.086 mile/h

Hence, speed of Tyson Gay's in mile/ hr is equal to 23.086 mph.

4 0

2 years ago

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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

2 years ago

Would u rather/ be nba young boy or polo g

denis-greek [22]

Answer:

nba young bruuhh

Explanation:

have a great day and plz mark brainliest!

6 0

2 years ago

Which statement is true of equinoxes? They occur in June and December. Days and nights are equal in length everywhere. The lengt

Zanzabum

<span>Days and nights are equal in length everywhere.(gradpoint)</span>

8 0

3 years ago

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Each step in any energy conversion process will _____.