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irga5000 [103]
3 years ago
7

An aurora occurs when _____.

Physics
2 answers:
drek231 [11]3 years ago
6 0
The correct answer is A. Charged particles from the sun exite the atmosphere near the poles to create auroras.
Bas_tet [7]3 years ago
6 0

Answer:

An aurora occurs when <u>the Earth's magnetic field is overwhelmed by charged ions.</u>

Explanation:

The Earth acts as a huge bar magnet. Magnetic field surrounds the Earth. when the charged particles coming from the sun interact with the magnetic field, beautiful colors can be seen in the sky. These are known as auroras. The color depends upon the gas present in the atmosphere. The electrons of gain energy and jump to higher level and then release photons when return back to its level.

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If a ball with an original velocity of zero is dropped from a tall structure and it takes 7 seconds to hit the ground, what velo
Colt1911 [192]
The velocity of the ball when it reaches the ground is equal to B. 68.6 m/s. This value was obtained from the formula Vf = Vi + at. Vf is the final velocity. Vi is the initial velocity. The acceleration is "a", while the time of travel is "t". The solution is:

<span>Vf = Vi + at
</span>Vf = 0 + (-9.8 m/s^2) (7 s)
Vf = -68.6 m/s

The negative sign denotes the direction of the ball.
5 0
3 years ago
Read 2 more answers
a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its
lilavasa [31]

Answer:

D/H =15

Explanation:

  • We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
  • At this point, we can find the value of H, applying the following kinematic equation:

       v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)

  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

       H = \frac{v_{0} ^{2} }{2*g} (2)

  • We can use the same equation, to find the value of D, as follows:

        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
  • When we replace these values in (1), we find that  v₁ = -v₀.
  • Replacing in (3), we have:

        (4*v_{0})^{2} - (-v_{0}) ^{2}  = 2* g* D\\ \\ 15*v_{0}^{2}  = 2*g*D

  • Solving for  D:

       D = \frac{15*v_{0} ^{2} }{2*g}

  • From (2) we know that H can be expressed as follows:

       H = \frac{v_{0} ^{2} }{2*g}

  • ⇒ D = 15 * H

        \frac{D}{H} = 15

3 0
3 years ago
Buco novia de 15n de edad
Zigmanuir [339]

Yo tengo quince y soy de cali y estoy viviendo en mexico

4 0
3 years ago
A 2kg bowling ball rolls at a speed of 5 m/s on a roof of the building that is 40 meters tall. What is the kinetic energy
MrRa [10]
The kinetic energy is \frac 1 2 m v^2 and the height of the building doesn't matter at all.

E = \frac 1 2 m v^2 = \frac 1 2 (2)(5)^2 = 25 joules
 
8 0
3 years ago
A 20-kg barrel is rolled up a 20-m ramp to the back of a truck whose floor is 5.0 m above the ground. What work is done in loadi
oee [108]

The angle of inclination is calculated using sin function,

sin θ = 5 m / 20 m = 0.25

θ = 14.4775° 

 

<span>The net force exerted is then calculated:
F net = m g sin θ = 20 * 9.8 * 0.25 </span>

F net = 49N 

<span>Work is product of net force and distance:
W = F net * d = 49 * 20 </span>

<span>Work = 980 J </span>

4 0
3 years ago
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