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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.

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Which statement best defines constructive interference? O Energy reflects back toward the source of its power. O Two waves with

Brums [2.3K]

Answer: two waves with identical crests and troughs meet

Explanation:

My teacher gave me the answer

4 0

2 years ago

A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats

ololo11 [35]

Answer:

Part a)

$f_B = 290 Hz$

Part B)

percentage increase is

$percentage = 1.38$ %

Explanation:

Part a)

As we know that the beat frequency is

$f_A - f_B = 3$

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

$293 - f_B = 3$

$f_B = 290 Hz$

Part B)

percentage increase in the tension of the string will be given as

$f_A - f_B' = 1$

$f_B' = 292 Hz$

now we have

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

so we have

$T_1 = C (290)^2$

$T_2 = C(292)^2$

so we have

$\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}$

percentage increase is

$percentage = 1.38$

4 0

3 years ago

Singly charged uranium-238 ions are accelerated through a potential difference of 2.90 kV and enter a uniform magnetic field of

Lera25 [3.4K]

Answer: 0.091 m

Explanation:

r = 1/B * √(2mV/e), where

r = radius of their circular path

B = magnitude of magnetic field = 1.29 T

m = mass of Uranium -238 ion = 238 * amu = 238 * 1.6*10^-27 kg

V = potential difference = 2.9 kV

e = charge of the Uranium -238 ion = 1.6*10^-19 C

r = 1/1.29 * √[(2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]

r = 1/1.29 * √(2.21*10^-21 / 1.6*10^-19)

r = 1/1.29 * √0.0138

r = 1/1.29 * 0.117

r = 0.091 m

Therefore, the radius of their circular path is 0.091 m

6 0

3 years ago

Calculate the frequency if the number of revolutions is 300 and the paired poles are 50.

Andrei [34K]

Answer: A

Explanation: We know that f=p*n

f=50*300=15000 Hz = 15kHz.

Have a great day! <3

3 0

1 year ago

Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at

beks73 [17]

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

ΔT = T₁ - T₂

20 = 180 - T₂

T₂ = 160°F

8 0

3 years ago