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Ghella [55]
3 years ago
10

Auroras occur at _____ of a planet. the equator the north magnetic pole only the south magnetic pole only both magnetic poles

Physics
1 answer:
goldenfox [79]3 years ago
3 0

<span>
The answer is - both magnetic poles</span>

<span>In the earth's southern magnetic pole, auroras are known as aurora austalis  or southern lights while in the northern magnetic pole they are known as aurora borealis or northern lights.</span>

<span>They are caused by charged particles borne in solar winds that are escaping from the sun. As they approach the earth,  the winds distort the earth's magnetic field which allows some particles to enter the earth's atmosphere, where they excite the gases and make them glow with spectacular display of colors.</span>

<span>
</span>

<span>
</span>

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From the top of the engineering building, you throw a ball vertically upward. the ball strikes the ground 4.00 s later. the engi
anastassius [24]
Equations of the vertical launch:

Vf = Vo - gt

y = yo + Vo*t - gt^2 / 2

Here yo = 35.0m
Vo is unknown
y final = 0
t = 4.00 s
and I will approximate g to 10m/s^2

=> 0 = 35.0 + Vo * 4 - 5 * (4.00)^2 => Vo = [-35 + 5*16] / 4 = - 45 / 4 = -11.25 m/s

The negative sign is due to the fact that the initial velocity is upwards and we assumed that the direction downwards was positive when used g = 10m/s^2.

Answer: 11.25 m/s


5 0
3 years ago
Read 2 more answers
To win a prize at the county fair, you're trying to knock down a heavy bowling pin by hitting it with a thrown object. Should yo
Setler [38]

Answer:

Being an elastic object, rubber ball will be an ideal choice as it will bounce off the bowling pit and will experience a large change in momentum in comparison with the beanbag which will either slow down or come to a halt upon hitting a bowling pit. That is why rubber ball will experience a greater impulse and the bowling pin will experience the negative impulse of the rubber ball.

For Rubber Ball

Upon elastic collision it will reverses the direction and move with velocity equal or less then original

change in momentum = P

P = m(v_{f} -v_{i})\\v_{f}=-v_{i} \\  P = -2mv_{i}

For Beanbag

value of impulse will large if velocity is zero.

v_{f}=0\\ P = -mv_{i}

Explanation:

8 0
3 years ago
A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart
serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

E_{initial} = E_{final}

PE_{initial}+KE_{initial} = PE_{final}+KE_{final}

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}

KE_{final} = (5100+4200)-5700

KE_{final} = 3600MJ

Therefore the final kinetic energy is 3600MJ

5 0
3 years ago
You are sitting on a deck of your house surrounded by oak trees. You hear the sound of an acorn hitting the deck. You wonder if
Black_prince [1.1K]

Answer: 96N

Explanation:

To calculate the velocity of the impact On the persons head, we have

h = gt²/2

14 = 9.81t²/2

t² = 28/9.8

t² = 2.86

t = 1.69s

V = u + at

V = 0 + 9.81*1.69

V = 16.58m/s

a(average) = (v1² + v2²) /2Δy

a(average) = 16.58² + 0)/2 * 0.005

a(average) = 274.8964/0.01

a(average) = 27489.64m/s²

Using newton's second law of motion,

F(average) = m * a(average)

F(average) = 0.0035 * 27489.64

F(average) = 96.21N

Therefore the force needed by the acorn to do much damage starts from 96N

8 0
3 years ago
Humans evolved in Earth's atmosphere, therefore the pressures interior to the body are relatively close to atmospheric pressure.
harina [27]

Answer:

194516 sheets

Explanation:

So the area of each sheet of paper is:

A = 0.216 * 0.279 = 0.060264 square meters

For the paper sheet to make the same effect as the atmospheric pressure P, then the gravity F from the paper sheet must be

F = AP = 0.060264 * 101325 = 6106 N

Let g = 9.81 m/s2, then the mass of paper needed to generate that gravity is

m = F/g = 6106 / 9.81 = 622.4 kg

If each sheet has a mass of 0.0032 kg, then the total number of sheets to have that much mass is

622.4 / 0.0032 = 194516 sheets

4 0
3 years ago
Read 2 more answers
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