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MA_775_DIABLO [31]

3 years ago

14

Where do the electrons that form the auroras enter the magnetosphere? a. Through holes c. At the equator b. Between the magnetic

field lines d. In the magnetotail

2 answers:

marta [7]3 years ago

7 0

I think the answer is d. In the magnetotail. I hope this helps! :)

sergiy2304 [10]3 years ago

3 0

Answer:

Between the magnetic field lines.

Explanation:

The electron which forms the auroras can enter the magnetosphere through the invisible magnetic field lines. These charge particles comes from the sun as in the form of wind.

Therefore, the electrons which forms the auroras can enter the magnetosphere in between the magnetic field lines which are invisible.

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It takes you 8.3 min to walk with an average velocity of 1.6 m/s to the north from the bus stop to the museum entrance. How far

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solution:

1.6 m/s = 96 m/min (in other words, 1.6 m/s x 60 s/min)

96 m/min x 8.3 min = 796.8 m

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Displacement vectors of 4 km north, 2km south, 5 km north, and 5km south combine to a total displacement of A 2 km south B 16 km

Phantasy [73]

The north vectors add up as so the south vectors. Then subtract the two. For north its 4 + 5 = 9. South is 2 + 5 = 7. Then 9-7 = 2km North (D)

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3 years ago

When an object is in a gravitational field, it has energy in its __________ __________ energy store.what’s is the word in the mi

lisabon 2012 [21]

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Explanation:

8 0

3 years ago

If a 6.1 A resistive load is connected by 2-conductor stranded 2-AWG copper wire to a source voltage of 125.2 V that is 358 ft a

polet [3.4K]

Answer:

124.86 V

Explanation:

We have to first calculate the voltage drop across the copper wire. The copper wire has a length of 358 ft

1 ft = 0.3048 m

358 ft = 109.12 m

The diameter of 2 AWG copper wire (d) = 6.544 mm = 0.006544 m

The area of the wire = πd²/4 = (π × 6.544²)/4 = 33.6 mm²

Resistivity of wire (ρ) = 0.0171 Ω.mm²/m

The resistance of the wire = $\frac{\rho A}{l}=\frac{0.0171*109.12 }{33.6} =0.056\ ohm$

The voltage drop across wire = current * resistance = 6.1 A * 0.056 ohm = 0.34 V

The voltage at end = 125.2 - 0.34 = 124.86 V

3 0

3 years ago

A student sits on a rotating stool holding two 3.09-kg masses. When his arms are extended horizontally, the masses are 1.08 m fr

schepotkina [342]

Answer:

a

The New angular speed is $w_f = 2.034 rad/s$

b

The Kinetic energy before the masses are pulled in is $KE_i = 3.101 \ J$

c

The Kinetic energy after the masses are pulled in is $KE_f = 8.192 \ J$

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

The radius $r =1.08m$

The mass is $m = 3.09\ kg$

The angular speed $w = 0.770 \ rad/sec$

The moment of inertia of the system excluding the two mass $I = 3.25 \ kg \cdot m^2$

New radius $r_{new} = 0.34m$

Generally the conservation of angular momentum can be mathematical represented as

$w_f = [\frac{I_i}{I_f} ]w_i .....(1)$

Where $w_f$ is the final angular speed

$w_i$ is the initial angular speed

$I_i$ is the initial moment of inertia

$I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

$I = m r^2$

Where I is the moment of inertia

m is the mass

r is the radius

So the Initial moment of inertia is given as

$I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

$I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

$I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final moment of inertia is given as

$I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$

Substituting these values into equation 1

$w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$

Generally Kinetic energy is mathematically represented in term of moment of inertia as

$KE = \frac{1}{2} * I * w^2$

Now considering the kinetic energy before the masses are pulled in,

$KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be $I_i = 10.46 \ Kg \cdot m^2$

The Angular speed would be $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

$KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

Now considering the kinetic energy after the masses are pulled in,

$KE_f = \frac{1}{2} * I_f * w^2_f$

The Moment of inertia would be $I_f = 3.96 \ Kg \cdot m^2$

The Angular speed would be $w_f = 2.034 \ rad/s$

Now substituting these value into the equation above

$KE_f= \frac{1}{2} *(3.96)(2.034)^2$

$= 8.192J$

8 0

3 years ago