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1 year ago

8

The dwarf planet praamzius is estimated to have a diameter of about 300km and orbits the sun at a distance of 6.4E12m . What is

it’s orbital period in years?

1 answer:

RoseWind [281]1 year ago

8 0

Answer:

Time period of the planet is

$T_2 = 278.7 years$

Explanation:

As we know by Kepler's law of time period

square of time period of planet is proportional to the cube of its orbital radius

$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

here we know for Earth

$T_1 = 1 year$

$R_1 = 1.5 \times 10^{11} m$

now we have

$\frac{1^2}{T_2^2} = \frac{(1.5 \times 10^{11})^3}{(6.4 \times 10^{12})^3}$

$T_2^2 = 7.77 \times 10^4$

$T_2 = 278.7 years$

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The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em> $1.83\times 10^{7}\,kg$ <em>. The wheels of the locomotive push back on the tracks with a constant net force of </em> $7.50\times 10^{5}\,N$ <em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force ( $F$ ), measured in newtons, then acceleration ( $a$ ), measured in meters per square second, must be constant and can be found by the following expression:

$a = \frac{F}{m}$ (1)

Where $m$ is the mass of the freight train, measured in kilograms.

If we know that $F = 7.50\times 10^{5}\,N$ and $m = 1.83\times 10^{7}\,kg$ , then the acceleration experimented by the train is:

$a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}$

$a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$

Now, the time taken to accelerate the freight train from rest ( $t$ ), measured in seconds, is determined by the following formula:

$t = \frac{v-v_{o}}{a}$ (2)

Where:

$v$ - Final speed of the train, measured in meters per second.

$v_{o}$ - Initial speed of the train, measured in meters per second.

If we know that $a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$ , $v_{o} = 0\,\frac{m}{s}$ and $v = 22.222\,\frac{m}{s}$ , the time taken by the freight train is:

$t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s} }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }$

$t = 542.265\,s$

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

6 0

1 year ago

A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity

Salsk061 [2.6K]

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

$\omega_i = 2\pi f_1$

$\omega_1 = 2\pi(\frac{610}{60})$

$\omega_1 = 63.88 rad/s$

similarly final angular speed is given as

$\omega_f = 2\pi f_2$

$\omega_2 = 2\pi(\frac{837}{60})$

$\omega_2 = 87.65 rad/s$

angular acceleration of the turbine is given as

$\alpha = 5.9 rad/s^2$

now we have to find the angular displacement is given as

$\theta = \omega t + \frac{1}{2}\alpha t^2$

$\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)$

$\theta = 234.62 radian$

3 0

1 year ago

An astronaut takes what he measures to be a 10-min nap in a space station orbiting Earth at 8000 m/s. A signal is sent from the

svet-max [94.6K]

Answer:

longer than

Explanation:

given,

time of nap = 10 min

speed of orbiting earth = 8000 m/s

c is the speed of light

using the equation of time dilation

$t' = \dfrac{t}{\sqrt{1-\dfrac{v^2}{c^2}}}$

now inserting all the values

$t' = \dfrac{10}{\sqrt{1-\dfrac{8000^2}{3\times 10^8)^2}}}$

$t' = \dfrac{10}{0.9999}$

t' = 10.001 s

on solving the above equation we will get a value greater than 10minutes.

hence, On earth time of nap measured will be longer than 10 min

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2 years ago

By constructing basic electric circuits, you will be able to measure current flow. With Ohm’s law in mind, what broad question a

andrezito [222]

Broad questions that we can answer by doing this experiment are <em><u>the effects of electric current</u></em>

<h3>Further explanation</h3>

Electric current is the amount of electric charge that flows each unit of time

Electric current occurs due to the movement of electrons due to the difference in potential or voltage (from high potential to low potential) between two points

Electrons will flow through the conducting wire that functions as a conductor

Ohm's Law states that

<em>The potential difference is proportional to the electric current as long as the resistance is constant</em>

$\displaystyle I=\frac{V}{R}\\\\V=I\times R$

A simple electrical circuit consists of a voltage source (battery) and a lamp

Ampermeters to measure the strength of the current, must be installed in series with the load to be measured

By changing the voltage source, with constant resistance from a conductor, different current measurements will be obtained. The greater the voltage, the greater the resulting current.

<h3>Learn more</h3>

electrons flow through the device

brainly.com/question/4438943

Keywords : basic electric circuits, Ohm’s law, experiment

5 0

1 year ago

Read 2 more answers

A technician is troubleshooting a problem. The technician tests the theory and determines the theory is confirmed. Which of the

vladimir1956 [14]

Answer:

b) Document lessons learned.

Explanation:

First he should do documentation

then C

then D

then A

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1 year ago