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polet [3.4K]
3 years ago
7

Which bright solar feature is shown in the picture above?

Physics
2 answers:
Ghella [55]3 years ago
6 0

Answer : (B) Prominence

Explanation :

A large, glittering and gaseous characteristic which is extending outward from the surface of the sun is called <em>Prominence</em>.

<em>Photosphere</em> is one of the layer of sun where the prominence are anchored and then they move into the corona of the sun.

<em>Corona</em> is a region in the surface of the sun which is the constituent of hot ionized gases (plasma).

The prominence consists of colder plasma and this prominence plasma is much more shining and denser as compared to coronal plasma.

Hence, the correct option is (B) Prominence.

bearhunter [10]3 years ago
3 0

The hottest spot in the feature, is the core, But in this situation it might be Prominence, B.)

Hope this helps!

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A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.
enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

a) Using the principle of energy conservation,

KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0
3 years ago
How do salt and water form a solution?
iogann1982 [59]
It would be D. The solute, salt, must dissolve IN the solvent, water (:
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Jane, looking for tarzan, is running at top speed (6.8 m/s) and grabs a vine hanging vertically from a tall tree in the jungle.
erica [24]
Jane's mechanical energy at any time is
E=U+K
where U=mgh is the potential energy, while K= \frac{1}{2} mv^2 is the kinetic energy.

Initially, Jane is on the ground, so the altitude is h=0 and the potential energy is zero: U=0. She's running with speed v, so she has kinetic energy only:
E=K= \frac{1}{2} mv^2
Then she grabs the vine, and when she reaches the maximum height h, her speed is zero: v=0, and so the kinetic energy becomes zero: K=0. So now her mechanical energy is just potential energy:
E=U=mgh

But E must be conserved, so the initial kinetic energy must be equal to the final potential energy:
\frac{1}{2}mv^2=mgh
from which we can find h, the maximum height Jane can reach:
h= \frac{v^2}{2g}= \frac{(6.8 m/s)^2}{2\cdot 9.81 m/s^2}=2.36 m
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what is the relationship between the evolution of marsupials and the movement of earths tectonic plate? 8th Grade No links or fi
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Steel is very stiff, and the Young's modulus for steel is unusually large, 2.0×1011 N/m2. A cube of steel 25 cm on a side suppor
TEA [102]

Answer:

Force (normal) = 833.85 N Compression = 1.67 x 10⁻⁸ m

Explanation:

Given data Young's Modulus (Y) = 2 x 10¹¹ N/m², Length of one side of cube = 25 cm = 0.25 m, mass of load = 85 kg

Normal force is the force exerted upon an object that is in contact with another stable object. This force would be applied by the surface onto the object in the same vector and is used to keep the object stable while it rests on a surface.

We know from Newton's Second Law that

F = ma where m is the mass and a is the acceleration (<em>in this case due to gravity</em>) hence, the normal opposing force to the load applied by the surface would be equal to the force applied on the surface by the weight of the load on the surface, So

F (normal) = M (load) x a = 85 x 9.81 = 833.85 N

Compression is the change in length of an object by the exertion of force upon it. Using the Young's Modulus formula we can find this change in the cube of steel. The Young's Modulus is given by

Y = (F/A)/(ΔL/L), where Y is the Young's Modulus, F is the Force being applied on the object, A is the cross sectional area on which the said force is applied, ΔL is the change in length due to said force being applied and L is the original Length of the side of the cross sectional area.

Solving this for ΔL, we can re- arrange the equation

ΔL = (F x L)/(Y x A) since area of square is L x L we can simplify the equation to get

ΔL = (F)/(Y x L), substitute the values

ΔL = (833.85)/(2 x 10¹¹ x 0.25) = 1.67 x 10⁻⁸m

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3 years ago
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