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torisob [31]
3 years ago
6

Future space stations will create an artificial gravity by rotating. Consider a cylindrical space station of 380 m diameter rota

ting about its axis. Astronauts walk on the inside surface of the space station. What rotation period will provide "normal" gravity? On a space walk on the outside of the space station how much gravity would they experience?
Physics
1 answer:
viktelen [127]3 years ago
8 0

Answer:

27.66 s

Explanation:

Space station creates artificial gravity by rotational movement about its axis .

The object inside also move in circular motion creating centrifugal force which creates acceleration in them .

centrifugal acceleration = ω² R where ω is angular velocity and R is radius of the cylindrical space station .

R = 380 /2 = 190 m

Given

ω² R = g = 9.8

ω² = 9.8 / R

= 9.8 / 190

= 5.15x 10⁻²

ω = 2.27 x 10⁻¹

= .227 rad / s

2π / T = .227 ( T is time period of rotation )

T = 2π / .227

= 27.66 s .

outside of the space station they will experience zero acceleration , because they are rotating around the earth.

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For what absolute value of the phase angle does a source deliver 71 % of the maximum possible power to an RLC circuit
kherson [118]

Answer: the absolute value of the phase angle is 28°

Explanation:

taking a look at expression for the instantaneous electric power in an AC circuit;

P = VI -------let this be equation 1

p is power, v is voltage and I is current;

for maximum power

P_max = V_rms × I_rms --------let this be equ 2

where P_max is the maximum power, V_rms is the rms value of voltage and I_rms is the rms value of current.

Also for average electric power in an AC circuit

P_avg = V_rms × I_rms × cos²∅ -------let this be equ 3

where P_avg is the average power and cos∅ is the power factor

now from equation 2;  P_max = V_rms × I_rms

so p_max replaces V_rms × I_rms in equation 3

we now have

P_avg = P_max × cos²∅

so we substitute

expression for the given value of the average power is

P_avg = P_max × 75%

p_avg = P_max.78/100

for the expression of the average electricity in an AC circuit

P_max.78/100 = P_max × cos²∅

78/100 = cos²∅

to get the absolute value of phase angle

∅ = cos⁻¹ ( √(78/100))

∅ =  cos⁻¹ ( 0.8832)

∅ = 27.969 ≈ 28°

Therefore the absolute value of the phase angle is 28°

8 0
3 years ago
List the types of electromagnetic radiation in order from lowest energy photons to highest energy photons.
frutty [35]

radio waves,X-rays,

Explanation:

In order from highest to lowest energy, the sections of the EM spectrum are named: gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, and radio waves. Microwaves (like the ones used in microwave ovens) are a subsection of the radio wave segment of the EM spectrum.

8 0
3 years ago
If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy
ruslelena [56]

The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

  • <em>distance of the galaxy, x = 2,000 Mpc</em>

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\  \times \ years}{3600 \ s \ \times\  24\ h\  \times \ 365.25 \ days} \\\\T = 1.397  \times 10^{10} \ years\\\\T = 13.97 \ billion \ years

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

8 0
3 years ago
Spacecrafts have landed on the surface of which planet?
ki77a [65]
I'm pretty sure its Venus!!!
7 0
3 years ago
A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti
vichka [17]

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

q(t) = e^{-\frac{t}{(R*C)}}

Here,

q = Charge

t = Time

R = Resistance

C = Capacitance

When the charge reach its half value it has passed 10ms, then the equation is,

\frac{1}{2}*q_{final} = e^{-\frac{0.01}{(R*C)}}

Ln(\frac{1}{2}) = -\frac{0.01}{RC}

- RC = \frac{0.01}{Ln(1/2)}

RC = 0.014s

We know that RC is equal to the time constant, then

T = RC  = 0.014s = 14ms

Therefore the time constant for the process is about 14ms

5 0
3 years ago
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