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3 years ago

Describe the two types of comet tails and how each are formed.

Physics

1 answer:

sesenic [268]3 years ago

6 0

There are two main types of cometary tails: the ion and dust tail.

NOTE: you can use quizlet, is helps out a lot

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An unlabeled hierarchical diagram of various astronomical bodies is shown below. The labels A, B, C, and D can be used to repres

Paha777 [63]

A) The biggest astronomical object is the Universe, which contains billions of galaxies among which there is the Milky Way.
The Milky Way contains thousands of planetary systems, among which the Solar System.
The Solar System contains many <span>planets <span>(but only one star, the Sun)</span>,</span> among which there is Earth.
Therefore you can label:
A = Universe, B = Milky Way, C = Solar system, D = Earth

b) Given what we said before, you could label D also any other planet in the Solar System, therefore you can choose among Mercury, Venus, Mars, Jupiter, Saturn, Uranus, and Neptune.

8 0

3 years ago

Wich of the following celestial bodies is most likely to have many craters

Ede4ka [16]

Where are the following awnsers?

6 0

3 years ago

An airplane accelerates down a runway at 3 m/s2 for 32 s until is finally lifts off the ground. Determine the distance traveled

Alex Ar [27]

Answer:

12 or 24

Explanation:

i think it is i hope it is right

4 0

3 years ago

Nitrogen at 100 kPa and 25oC in a rigid vessel is heated until its pressure is 300 kPa. Calculate (a) the work done and (b) the

nignag [31]

Answer:

A. The work done during the process is W = 0

B. The value of heat transfer during the process Q = 442.83 $\frac{KJ}{kg}$

Explanation:

Given Data

Initial pressure $P_{1}$ = 100 k pa

Initial temperature $T_{1}$ = 25 degree Celsius = 298 Kelvin

Final pressure $P_{2}$ = 300 k pa

Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.

⇒ $V_{1}$ = $V_{2}$ ------------- (1)

Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.

⇒ P ∝ T

⇒ $\frac{P_{2} }{P_{1}}$ = $\frac{T_{2} }{T_{1}}$

⇒ Put all the values in the above formula we get the final temperature

⇒ $T_{2}$ = $\frac{300}{100}$ × 298

⇒ $T_{2}$ = 894 Kelvin

(A). Work done during the process is given by W = P × ( $V_{2} -V _{1}$ )

From equation (1), $V_{1}$ = $V_{2}$ so work done W = P × 0 = 0

⇒ W = 0

Therefore the work done during the process is zero.

Heat transfer during the process is given by the formula Q = m $C_{v}$ ( $T_{2}$ - $T_{1}$ )

Where m = mass of the gas = 1 kg

$C_{v}$ = specific heat at constant volume of nitrogen = 0.743 $\frac{KJ}{kg k}$

Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )

⇒ Q = 442.83 $\frac{KJ}{kg}$

Therefore the value of heat transfer during the process Q = 442.83 $\frac{KJ}{kg}$

6 0

3 years ago

The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular

nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

6 0

3 years ago