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4 years ago

The cloud of interstellar dust and gas that forms a star is known as a

1 answer:

3 0

D. nebula

Nebula is the cloud of interstellar dust and gas that forms a star. It is collborately shaped during colliding and collapsing of the interstellar mediums this is influenced by the gravitational attraction of the atoms and particles in the entites. Hence, there are three types of nebular namely, are classical nebula, diffuse nebula, planetary nebular and supernova remnants.

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URGENT!!!!! I WILL MARK U AS BRAINLIEST!!!!!!! A step-down transformer converts 120 V to 5 V in a phone charger. If the primary

Sloan [31]

Answer:

25 turns

Explanation:

The voltage ratio is the turns ratio:

secondary turns / primary turns = secondary voltage / primary voltage

secondary turns / 600 = 5/120

secondary turns = 600(5/120) = 25 . . . turns

The secondary coil has 25 turns.

6 0

3 years ago

Which will speed up a chemical reaction? Apply heat. Increase the concentration of the reactants. Grind up the substance, so the

makkiz [27]

All of the above options are correct

8 0

3 years ago

In the circuit diagram, what does the symbol made of two Long lines and two short lines with a positive and a negative sign at e

Svet_ta [14]

Yeah it is a source of electric energy

7 0

3 years ago

Read 2 more answers

Two 2.0-cm-diameter insulating spheres have a 6.60 cm space between them. One sphere is charged to + 76.0 nC , the other to - 30

e-lub [12.9K]

Answer:

$5.2\times 10^5N/C$

Explanation:

Since the two charged bodies are symmetric, we can calculate the electric field taking both of them as point charges.

This can be easily seen if we use Gauss's law, $\int{E} \, dA=\frac{Q_{enclosed}}{\epsilon_o}$

We take a larger sphere of radius, say r, as the Gaussian surface. Then the electric field due to the charged sphere at a distance r from it's center is given by,

$E=\frac{1}{4\pi r^2} \frac{Q_{enclosed}}{\epsilon_o}$

which is the same as that of a point charge.

In our problem the charges being of opposite signs, the electric field will add up. Therefore,

$E_{total}=\frac{1}{4\pi\epsilon_o}\frac{q_1+q_2}{r^2}= (9\times10^9) \frac{(76+30)\times10^{-9}}{((1+3.3)\times10^{-2})^2}N/C =5.2\times10^5N/C$

where, $r$ = distance between the center of one sphere to the midpoint (between the 2 spheres)

8 0

3 years ago

Assuming this is a distance time graph( ignore the speed time title) assume metres on vertical scale. describe in as much detail

riadik2000 [5.3K]

Answer:

The journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance travelled during the journey from the start point A to the final point B is 40 m

Explanation:

From the start point A to point B, we have;

The speed from A to B = 10 m/(10 s) = 1 m/s

The distance traveled from A to B = 10 m

The time it takes to move from A to B = 10 seconds

From the point B to point C, we have;

The distance traveled from B to C = 0 m, (stationary)

The time it remains at point B distance from the start point = 10 seconds

The speed between point B to C = 0 m/(10 s) = 0 m/s

From the point C to point D, we have;

The distance traveled from C to D = 10 m

The time it takes to move from C to D = 5 seconds

The speed between point C and D = 10 m/(5 s) = 2 m/s

From the point D to point E, we have;

The distance traveled from D to E = 0 m, (stationary)

The time it remains at point D distance from the start point = 10 seconds

The speed between point D to E = 0 m/(10 s) = 0 m/s

From the point E to point F, we have;

The distance traveled from E to F = 20 m (return journey starts at point E)

The time it takes to move from E to F = 5 seconds

The speed between point E to F = 20 m/(5 s) = 4 m/s (Return journey)

Therefore, the journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance moved, 'd', to and from the start point with reference to the graph is given as follows;

d = (From A to B) 10 m + (From B to C) 0 m + (From C to D) 10 m + (From D to E) 0 m + (From E to F) 20 m = 40 m

The total distance travelled in the journey is 40 m

The total displacement, $\underset{d}{\rightarrow}$ = 10 m + 10 m - 20 m = 0 m

7 0

3 years ago