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2 years ago

Which of these diagrams best represents the steps in the formation of the sun?

1 answer:

4 0

The diagram best represents the steps in the formation of the sun is the clouds of gases condense > The clouds rotate at high speed > The clouds flatten into an elliptical disc > The planets are born. The correct option is D.

<h3>What are planets?</h3>

Planets are the large spherical shaped objects that rotate about the Sun in the elliptical orbits.

Planets are shaped from Planetary cloud. The dust storm and gases gathers under its own weight. The dense matter beginnings pivoting at high paces and accumulates more mass. The center structures, the star and rest of it ultimately levels into a curved plate from which planet is formed.

Thus, the correct sequence is D.

Learn more about planets.

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You might be interested in

Tyson Gay's best time to run 100.0 meters was 9.69 seconds. What was his average speed during this run, in miles per hour? (3.28

FinnZ [79.3K]

Answer:

23.086 mile/h

Explanation:

Given,

Distance Tyson Gay run = 100 m

time of run, t = 9.69 s

average speed of the in mph = ?

Speed of the Gay = $\dfrac{distance}{time}$

$v = \dfrac{100}{9.69}$

v = 10.32 m/s

1 m = 3.281 ft

10.32 m = 33.86 ft

1 mile = 5280 ft

1 ft = 1.8939 x 10⁻⁴ mile

33.86 ft/s = 6.413 x 10⁻³ miles/s

Speed of Tyson in mile/hr = 6.413 x 10⁻³ x 3600

= 23.086 mile/h

Hence, speed of Tyson Gay's in mile/ hr is equal to 23.086 mph.

4 0

3 years ago

Read 2 more answers

How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

$T_{f}$ = (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

$T_{f}$ = 466.8 K

so change in temperature ΔT

ΔT = 466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x $_{i\\}$ ² - x $_{f\\}$ ² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q = 16 J

Therefore, the required heat energy is 16 J

5 0

3 years ago

Two ranger stations are on an east-west line 110 mi apart. A forest fire is located on a bearing of N 42º E from the western sta

olchik [2.2K]

Answer:

Explanation:

AB = 110 miles

Let the distance of the western station from fire is d.

As according to the diagram, use Sine law

$\frac{d}{Sin 15}=\frac{110}{Sin 133}$

d = 110 x 0.2588 / 0.73

d = 39 miles

8 0

3 years ago

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

3 years ago

A 6.0 m wire with a mass of 50 g, is under tension. A transverse wave, for which the frequency is 810 Hz, the wavelength is 0.40

MrRissso [65]

Answer:

a) t = 0.0185 s = 18.5 ms

b) T = 874.8 N

Explanation:

First we find the seed of wave:

v = fλ

where,

v = speed of wave

f = frequency = 810 Hz

λ = wavelength = 0.4 m

Therefore,

v = (810 Hz)(0.4 m)

v = 324 m/s

Now,

v = L/t

where,

L = length of wire = 6 m

t = time taken by wave to travel length of wire

Therefore,

324 m/s = 6 m/t

t = (6 m)/(324 m/s)

<u></u>

From the formula of fundamental frquency, we know that:

Fundamental Frequency = v/2L = (1/2L)(√T/μ)

v = √(T/μ)

where,

T = tension in string

μ = linear mass density of wire = m/L = 0.05 kg/6 m = 8.33 x 10⁻³ k gm⁻¹

Therefore,

324 m/s = √(T/8.33 x 10⁻³ k gm⁻¹)

(324 m/s)² = T/8.33 x 10⁻³ k gm⁻¹

8 0

3 years ago