A friend tells you that a lunar eclipse will take place the following week, and invites you to join him to observe the eclipse through a high-powered telescope he owns. You are curious what the eclipse might look like from different perspectives in space. If the moon has a diameter of 2,159.14 miles, what is the maximum distance that it could be observed by the naked eye with enough detail that you could distinguish it from other celestial bodies (assuming that you have 20/20 vision)

Answers

Answer 1

Answer:

Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

θ = 1.22 550 10⁻⁹ / 0.002

θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

tan θ = y / L

y = L tan θ

y = 2,389 10⁵ tan 3,355 10⁻⁴

y = 8.02 10¹ mi

y = 80.2 mille

This is the smallest size of an object seen directly by the eye

Answer 2

Answer:

Final answer:

An individual with 20/20 vision can observe the moon from a maximum distance of around 6200 km or 3850 miles. Beyond this distance, it might be difficult to distinguish the moon from other celestial objects without using a telescope. The use of a telescope can expand this range significantly.

Explanation:

The detailed observation of a lunar eclipsed, when viewed without any form of optical aid like a telescope, is contingent on many factors, one of which is the human eye's angular resolution—the eye's ability to differentiate between two separate points of light. For an average human eye with 20/20 vision, the angular resolution is approximately 0.02 degrees.

To calculate the maximum distance at which the moon could be observed clearly with the eye, the formula for small angle approximation can be used, which in this context is: Distance = Size / Angle = (2159.14 miles) / (0.02 degrees in radians). This calculates to a distance of approximately 6200 km or 3850 miles.

Beyond this distance, distinguishing the moon from other celestial bodies might be challenging using just the eye. Utilizing a high-powered telescope would significantly extend this range by magnifying the image, allowing clearer detail over much greater distances.

Learn more about Observing Moon here:

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Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the compressor if the inlet volumetric flow rate is 0.15 m3/s. Use constant specific heats evaluated at 300 K.

Answers

Answer:

$\dot W_(in) = 49.386\,kW$

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

$\dot W_(in) + \dot m \cdot c_(p)\cdot (T_(1)-T_(2)) = 0$

The power consumed by the compressor can be calculated by the following expression:

$\dot W_(in) = \dot m \cdot c_(v)\cdot (T_(2)-T_(1))$

Let consider that air behaves ideally. The density of air at inlet is:

$P\cdot V = n\cdot R_(u)\cdot T$

$P\cdot V = (m)/(M)\cdot R_(u)\cdot T$

$\rho = (P\cdot M)/(R_(u)\cdot T)$

$\rho = ((104\,kPa)\cdot (28.02\,(kg)/(kmol)))/((8.315\,(kPa\cdot m^(3))/(kmol\cdot K) )\cdot (292\,K))$

$\rho = 1.2\,(kg)/(m^(3))$

The mass flow through compressor is:

$\dot m = \rho \cdot \dot V$

$\dot m = (1.2\,(kg)/(m^(3)))\cdot (0.15\,(m^(3))/(s) )$

$\dot m = 0.18\,(kg)/(s)$

The work input is:

$\dot W_(in) = (0.18\,(kg)/(s) )\cdot (1.005\,(kJ)/(kg\cdot K))\cdot (565\,K-292\,K)$

$\dot W_(in) = 49.386\,kW$

An 800-kHz sinusoidal radio signal is detected at a point 6.6 km from the transmitter tower. The electric field amplitude of the signal at that point is 0.780 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the amplitude of the magnetic field of the signal at that point

Answers

Answer:

0.0000000026 T

Explanation:

$E_0$ = Maximum electric field strength = 0.78 V/m

$B_0$ = Maximum magnetic field strength

c = Speed of light = $3* 10^8\ m/s$

Relation between amplitudes of electric and magnetic fields is given by

$E_0=B_0c\n\Rightarrow B_0=(E_0)/(c)\n\Rightarrow B_0=(0.78)/(3* 10^8)\n\Rightarrow B_0=0.0000000026\ T$

The amplitude of the magnetic field is 0.0000000026 T

The radius of a typical human eardrum is about 4.15 mm. Calculate the energy per second received by an eardrum when it listens to sound that is at the threshold of hearing, assumed to be 1.20E-12 W/m2

Answers

The energy per second received by an eardrum is $6.4884 * 10^(-17) watt$

Calculation of the energy per second;

The area should be

$= \pi r^2\n\n= 3.14 * 0.00415m\n\n= 5.407 * 10^(-5)m^2$

Now

The power should be

$= 1.2 * 10^(-12) * 5.407 * 10^(-5)\n\n= 6.4884 * 10^(-17) watt$

Learn more about the energy here: brainly.com/question/14338287

Answer:

Power energy per second will be equal to $6.4884* 10^(-17)watt$

Explanation:

We have given radius of human eardrum r = 4.15 mm = 0.00415 m

Intensity at threshold of hearing $I=1.2* 10^(-12)w/m^2$

Area is given by $A=\pi r^2=3.14* 0.00415^2=5.407* 10^(-5)m^2$

We know that power is given by $P=I* A=1.2* 10^(-12)* 5.407* 10^(-5)=6.4884* 10^(-17)watt$

So power energy per second will be equal to $6.4884* 10^(-17)watt$

Why does an astronaut in a spacecraft orbiting Earthexperience
a feeling of weightlessness?

Answers

Answer:

Astronaut in spacecraft while orbiting earth experience weightlessness because there is no gravity of earth or moon is acting on the body of an astronaut.

while on earth, we experience weight because the gravity of earth is acting on our body which is pulling us downward.

Both spacecraft and the astronauts both are in a free-fall condition.

What is displacement?

Answers

Displacement means when you move something from its original position. Let's say you want to sit on a chair. You move the chair from where it was originally placed. That's displacement.

Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.2 s and was brought jarringly back to rest in only 1 s. Calculate his (a) magnitude of acceleration in his direction of motion and (b) magnitude of acceleration opposite to his direction of motion. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity. g g

Answers

Answer:

a = 5.53 g , a = -15g

Explanation:

This is an exercise in kinematics.

a) Let's look for the acceleration

as part of rest v₀ = 0

v = v₀ + a t

a = v / t

a = 282 / 5.2

a = 54.23 m / s²

in relation to the acceleration of gravity

a / g = 54.23 / 9.8

a = 5.53 g

b) let's look at the acceleration to stop

va = 0

0 = v₀ -2 a y

a = vi / y

a = 282/2 1

a = 141 m /s²

a / G = 141 / 9.8

a = -15g

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