All categories

3 years ago

Stellar evolution is _____. the process of planetary formation the life cycle of a star the death of a star how comets form

Physics

2 answers:

frosja888 [35]3 years ago

8 0

B.) Stellar evolution is "<span>the life cycle of a star"

Hope this helps!</span>

Mnenie [13.5K]3 years ago

6 0

The life cycle of a star.
Planetary formation has nothing to do with the word stellar
The death of a star is only one phase in it's evolution
How comets form is a branch of planetary, as comets are actually speeding asteroids, or so I've heard :)

You might be interested in

Which is an example of a screw?

NikAS [45]

Which one looks like a screw
A is shoes
B is a fork (lever)
C is a wind chime shaped like a screw
D is a stapler (lever)

5 0

3 years ago

George determines the mass of his evaporating dish to be 3.375 g. He adds a solid sample to the evaporating dish, and the mass o

zhannawk [14.2K]

Explanation:

The given data is as follows.

Mass of evaporating dish = 3.375 g

Total mass = Mass of solid sample + evaporating dish

That is, Mass of solid sample + evaporating dish = 26.719 g

Therefore, we will calculate the mass of solid sample as follows.

Mass of solid sample = (Mass of solid sample + evaporating dish) - mass of evaporating dish

= 26.719 g – 3.375 g

= 23.344 g

Thus, we can conclude that mass of his solid sample must be 23.344 g.

3 0

4 years ago

A spherical balloon has a radius of 6.95m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of ai

klio [65]

Answer:

602.27 kg

Explanation:

The computation of the largest mass of cargo the balloon can lift is shown below:-

Volume of helium inside the ballon= (4 ÷ 3) × π × r^3

= (4 ÷ 3) × 3.14 × 6.953

= 1406.19 m3

Mass the balloon can carry = volume × (density of air-density of helium)

= 1406.19 × (1.29-0.179)

= 1562.27 kg

Mass of cargo it can carry = Mass it can carry - Mass of structure

= 1562.27 - 960

= 602.27 kg

6 0

3 years ago

After touchdown a fighter jet passes a marker on the runway at an instantaneous speed of 100 m/s and constant negative accelerat

daser333 [38]

Given that,

Deacelectration = -15 m/s²

Negative sign shows the declaration

Slowly speed = 25 m/s

Acceleration = 35 m/s²

Speed = 170 m/s

We need to calculate the time

Using equation of motion

$v=u+at$

Where, v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the formula

$25=0+15\times t$

$t=\dfrac{25}{15}$

$t=1.66\ sec$

We need to calculate the distance from the original runway

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}\times35\times(1.66)^2$

$s=48.2\ m$

Hence, The distance from the original runway is 48.2 m.

3 0

4 years ago

A playground merry-go-round with a radius of 1.80 m has a mass of 120 kg and is rotating with an angular speed of 0.350 rev/s. W

stepan [7]

Answer:

1.64 rad/s

Explanation:

Given,

radius of merry-go-round, r = 1.80 m

mass of merry-go-round, M = 120 Kg

angular speed, ω = 0.350 rev/s

Initial angular speed = 2π x 0.350 = 2.12 rad/s

Mass of child, m = 36.5 Kg

Moment of inertia of the merry-go-round

$I = \dfrac{1}{2}Mr^2$

$I = \dfrac{1}{2}\times 120\times 1.8^2$

$I = 349.92\ kg.m^2$

L= I ω = 349.92 x 2.12 = 769.41 kgm²/s

Child moment of inertia

I₂ = m r^2 = 36.5 x 1.8² = 118.26 kg.m²

Final moment of inertia,

I_f = 349.92+118.26 = 468.18 kg.m²

Final speed = $\dfrac{769.41}{468.14}$

$\omega_f = 1.64 \rad/s$

angular speed of the child is equal to 1.64 rad/s

4 0

4 years ago