Physics, 11.10.2020 23:01, miyahhh123

Determine the ratio β = v/c for each of the following. (a) A car traveling 120 km/h.
(b) A commercial jet airliner traveling 270 m/s.
(c) A supersonic airplane traveling mach 2.7. (Mach number = v/vsound. Assume the speed of sound is 343 m/s.)
(d) The space shuttle, traveling 27,000 km/h.
(e) An electron traveling 30 cm in 2 ns.
(f) A proton traveling across a nucleus (10-14 m) in 0.38 ✕ 10-22 s.

Answers: 1

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Answers

a) $\beta = 1.111\times 10^{-7}$ , b) $\beta = 9\times 10^{-7}$ , c) $\beta = 3.087\times 10^{-6}$ , d) $\beta = 2.5\times 10^{-5}$ , e) $\beta = 0.5$ , f) $\beta = 0.877$

Explanation:

From relativist physics we know that is the symbol for the speed of light, which equal to approximately 300000 kilometers per second. (300000000 meters per second).

a) A car traveling 120 kilometers per hour:

At first we convert the car speed into meters per second:

$v = \left(120\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v = 33.333\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 33.333\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{33.333\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 1.111\times 10^{-7}$

b) A commercial jet airliner traveling 270 meters per second:

The ratio $\beta$ is now calculated: ( $v = 270\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{270\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 9\times 10^{-7}$

c) A supersonic airplane traveling Mach 2.7:

At first we get the speed of the supersonic airplane from Mach's formula:

$v = Ma\cdot v_{s}$

Where:

- Mach number, dimensionless.

$v_{s}$ - Speed of sound in air, measured in meters per second.

If we know that Ma = 2.7 and $v_{s} = 343\,\frac{m}{s}$ , then the speed of the supersonic airplane is:

$v = 2.7\cdot \left(343\,\frac{m}{s} \right)$

$v = 926.1\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 926.1\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{926.1\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 3.087\times 10^{-6}$

d) The space shuttle, travelling 27000 kilometers per hour:

At first we convert the space shuttle speed into meters per second:

$v = \left(27000\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v = 7500\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 7500\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{7500\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 2.5\times 10^{-5}$

e) An electron traveling 30 centimeters in 2 nanoseconds:

If we assume that electron travels at constant velocity, then speed is obtained as follows:

$v = \frac{d}{t}$

Where:

- Speed, measured in meters per second.

- Travelled distance, measured in meters.

- Time, measured in seconds.

If we know that $d = 0.3\,m$ and $t = 2\times 10^{-9}\,s$ , then speed of the electron is:

$v = \frac{0.3\,m}{2\times 10^{-9}\,s}$

$v = 1.50\times 10^{8}\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 1.5\times 10^{8}\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{1.5\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 0.5$

f) A proton traveling across a nucleus (10⁻¹⁴ meters) in 0.38 × 10⁻²² seconds:

If we assume that proton travels at constant velocity, then speed is obtained as follows:

$v = \frac{d}{t}$

Where:

- Speed, measured in meters per second.

- Travelled distance, measured in meters.

- Time, measured in seconds.

If we know that $d = 10^{-14}\,m$ and $t = 0.38\times 10^{-22}\,s$ , then speed of the electron is:

$v = \frac{10^{-14}\,m}{0.38\times 10^{-22}\,s}$

$v = 2.632\times 10^{8}\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 2.632\times 10^{8}\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{2.632\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 0.877$

answered: Guest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

answered: Guest

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the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia