All categories

3 years ago

Why were the rings of Uranus not observed directly from telescopes on the ground on Earth? How were they discovered?

Physics

1 answer:

leonid [27]3 years ago

6 0

Answer:Explained below.

Explanation:

Uranus rings is made up of jet black, coal-like particles in small bands, making them difficult to perceive from Earth.This indicates that they are probably composed of a mixture of the ice and a dark material. The nature of material is dismal, but it might be some organic compounds greatly darkened by the charged particle irradiation from the Uranian magnetosphere. Rings were discovered by using a infrared telescope throughout the occultation of a star as Uranus passed in front of it. The light from the star dimmed many times before it was obstructed by the disk of Uranus and subsequently, showing the presence of various distinct rings.

You might be interested in

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

4 0

3 years ago

About once every 30 minutes, a geyser known as Old Faceful projects water 11.0 m straight up into the air. Use g = 9.80 m/s^2, a

Maurinko [17]

Answer:

The speed of the water is 14.68 m/s.

Explanation:

Given that,

Time = 30 minutes

Distance = 11.0 m

Pressure = 101.3 kPa

Density of water = 1000 kg/m³

We need to calculate the speed of the water

Using equation of motion

$v^2=u^2+2gs$

Where, u = speed of water

g = acceleration due to gravity

h = height

Put the value into the formula

$0=u^2-2\times9.8\times11.0$

$u=\sqrt{2\times9.8\times11.0}$

$u=14.68\ m/s$

Hence, The speed of the water is 14.68 m/s.

7 0

3 years ago

a ball is shot from the ground straight up into the air with initial velocity of 50 50 ft/sec. assuming that the air resistance

Anna [14]

The height attained by the ball is 11.86m

a ball is shot from the ground straight up into the air its initial and final velocity is

initial velocity, u = 50 ft/s = 50×0.305 = 15.25m/s

final velocity ,v = 0 m/s

gravity =-9.8 m/s²

( negative sign shows acceleration in opposite direction)

height =?

using the newton motion of equation

v² = u² + 2as

where

a= acceleration due to gravity(g)

s = height

v² = u² + 2gs

(0)² = (15.25)² + 2×(-9.8)×s

0 = (15.25)² - 19.6 × s

s= - (15.25)²/ 19.6

s = 11.86m

after ignoring the air resistance the maximum height of the ball is 11.86m

To learn more about motion under gravity -

brainly.com/question/27962354

#SPJ4

3 0

1 year ago

In a compound, the atom does or does not take on a new set of properties

disa [49]

It does take on new set of proerties

7 0

3 years ago

ENERGY SAVERS RACE, BRAIN BURNER. This question is about solar cars at the Chuck Norris Institute of Technology, CNIT, in Ocala,

Hunter-Best [27]

Answer:

a) d = 6.0 m

Explanation:

Since car is accelerating at uniform rate then here we can say that the distance moved by the car with uniform acceleration is given as

$d = \frac{(v_f + v_i)}{2} \times t$

here we know that

$v_f = 10 m/s$

$v_i = 0$

$t = 1.2 s$

now we will have

$d = \frac{(10 + 0)}{2}\times 1.2$

$d = 5 \times 1.2$

$d = 6.0 m$

4 0

3 years ago