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2 years ago

14

Future space stations could create an artificial gravity by rotating. Consider a cylindrical space station that rotates with a p

eriod of 28 seconds about its axis. Astronauts walk on the inside surface of the space station, and they experience an Earth-like gravitational acceleration of 9.8 m/s2. What is the diameter of the space station?
510m
145m
660m
390m

1 answer:

aleksandrvk [35]2 years ago

8 0

Answer:

P = 2 pi R / v period of space station

F / m = v^2 / R centripetal force per unit of mass

So F / m = 4 pi^2 R^2 / (P^2 * R) = 4 pi^2 R / P^2

Also, F / m = 9.8 m/s^2 earth's gravitational attraction

So 9.8 = 4 pi^2 R / P^2 or R = 9.8 P^2 / 4 * pi^2) = 195 m

Or D = 2 R = 390 m the diameter required

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A truck driver is broadcasting at a frequency of 27.075 MHz with a CB (citizen's band) radio. Determine the wavelength of the el

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Answer:

11m

Explanation:

Wavelength of a wave is the distance between successive crests and trough of a wave. It is represented mathematically as the ratio of the speed of the wave to its frequency. It can be expressed as;

Wavelength ¶ = wave speed (v) /frequency (f)

Given the speed of light c = 2.9979 10^8 m/s.

frequency of the radio wave = 27.075 MHz = 27.075×10^6Hz

Wavelength = 2.9979 10^8/27.075×10^6

Wavelength = 0.11×10^2

Wavelength = 11m

Therefore, the wavelength of the electromagnetic wave being used is 11m

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3 years ago

a lorry travels 3600m on a test track accelerating constantly at 3m/s squared from standstill. what is the final velocity (3 sig

suter [353]

The final velocity of the truck is found as 146.969 m/s.

Explanation:

As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.

Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.

$2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}$

Then, the final velocity will be

$v^{2} = 2 * 3 * 3600 + 0 = 21600\\ \\v=\sqrt{21600}=146.969 m/s$

Thus, the final velocity of the truck is found as 146.969 m/s.

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3 years ago

Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of 190

julia-pushkina [17]

Answer:

$5.5\cdot 10^{-11} C$

Explanation:

The capacitance of the parallel-plate capacitor is given by:

$C=\epsilon_0 \frac{A}{d}$

where

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$A=190 mm^2 = 190 \cdot 10^{-6} m^2$ is the area of the plates

$d=1.2 mm = 0.0012 m$ is the separation between the plates

Substituting,

$C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F$

The energy stored in the capacitor is given by

$U=\frac{Q^2}{2C}$

Since we know the energy

$U=1.1 nJ = 1.1 \cdot 10^{-9} J$

we can re-arrange the formula to find the charge, Q:

$Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C$

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2 years ago

The following is current scientific evidence supporting the nebular theory on the formation of the solar system. the composition

andreyandreev [35.5K]

All planets orbit the sun in a plane, all the planets orbit the sun in the same direction, most of the planets rotate in the same direction. I'm not sure when and answer ends or begins on your question so you can choose from some of the answers I gave you.

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3 years ago

Read 2 more answers

A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit

djverab [1.8K]

Answer:

The current is $I = 6.68 \ A$

Explanation:

From the question we are told that

The radius of the loop is $r = 6 \ cm = 0.06 \ m$

The earth's magnetic field is $B_e = 0.7G= 0.7 G * \frac{1*10^{-4} T}{1 G} = 0.7 *10^{-4} T$

The number of turns is $N =1$

Generally the magnetic field generated by the current in the loop is mathematically represented as

$B = \frac{\mu_o * N * I}{2 r }$

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

$B = B_e$

=> $B_e = \frac{\mu_o * N * I }{ 2 * r}$

Where $\mu$ is the permeability of free space with value $\mu _o = 4\pi * 10^{-7} N/A^2$

$0.7 *10^{-4}= \frac{ 4\pi * 10^{-7} * 1 * I}{2 * 0.06}$

=> $I = \frac{2 * 0.06 * 0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}$

$I = 6.68 \ A$

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3 years ago