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3 years ago

How are Aurora produced?

Physics

1 answer:

Juli2301 [7.4K]3 years ago

3 0

Answer:

The third one.

Explanation:

Atmospheric molecules releasing photons (light) in response to collisions with Solar Wind particles because when auroras are produced what actually happens is that the ions of the solar wind collide with atoms of oxygen and nitrogen from the Earth's atmosphere. The energy released during these collisions causes a colorful glowing halo around the poles which is the aurora.

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A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0

viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
$KE = \frac{1}{2}mv^2 \\\\PE = mgh$

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

$E_i = \frac{1}{2}mv_i^2 + mg(H_1 - H_2)$

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

$E_f = \frac{1}{2}mv_f^2$

And:
$W_A = E_i - E_f$

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.

Therefore:
$W_A = \frac{1}{2}mv_i^2 + mg(H_1 - H_2) - \frac{1}{2}mv_f^2$

Solving for the work done by air resistance:
$W_A = \frac{1}{2}(.450)(15.1^2)+ (.450)(9.8)(30.2 - 12) - \frac{1}{2}(.450)(19.89^2)$

$W_A = \boxed{42.552 J}$

8 0

2 years ago

A dog has a mass of 20 kg. If the dog is pushed across the ice with a force of 40 N, what is its acceleration?

olasank [31]

Answer:

The acceleration is 2 m/s2.

Explanation:

We calculate the acceleration (a), with the data of mass (m) and force (F), through the formula:

F = m x a ---> a= F/m

a = 40 N/20 kg <em> 1N= 1 kg x m/s2</em>

a= 40 kgx m/s2/ 20 kg

7 0

3 years ago

A 75.0 kg person is on a Ferris Wheel that has a radius R = 16 m. The tangential velocity of the Ferris Wheel is 8.25 m/s. Calcu

Dovator [93]

Answer:

The period of rotation is

T=8.025s

Explanation:

The person is undergoing simple harmonic motion on the wheel

Given data

mass of the person =75kg

Radius of wheel r=16m

Velocity =8.25m/s

The oscillating period of simple harmonic motion is given as

T=(2*pi)/2=2*pi √r/g

Assuming that g=9.81m/s

Substituting our data into the expression we have

T=2*3.142 √ 16/9.81

T=6.284*1.277

T=8.025s

3 0

3 years ago

Thalia is drafting a plan to move a large, perfect sphere concrete sculpture that is in front of her office building. Describe t

Charra [1.4K]

Answer:She would need to first know the weight of the sculpture and what she is going to move it with then she will need to use newton's second law to calculate the amount of force needed to move it

Explanation: I just did the assignment on edgunity

3 0

3 years ago

Read 2 more answers

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago