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Arturiano [62]
3 years ago
12

Near Earth, the density of protons in the solar wind (a stream of particles from the Sun) is 12.5 cm-3, and their speed is 564 k

m/s. (a) Find the current density of these protons. (b) If Earth's magnetic field did not deflect the protons, what total current Earth receive?
Physics
1 answer:
morpeh [17]3 years ago
7 0

Answer:

A) 11.28 x 10^(7) A.m²

B) 2.258 x 10^(17)A

Explanation:

A) The current density is given by the formula ;

J = nqv

Where n is the density of protons in the solar wind which is 12.5 cm³ or 12.5 x 10^(-6) m³

q is the proton charge which is 1.6 x 10^(-19) C

v is velocity which is 564km or 564000m

Thus, J = 12.5 x 10^(-6) x 1.6 x 10^(-19) x 564000 = 11.28 x 10^(7) A.m²

B) the formula for the total current the earth received is given as;

I = JA

The effective area is the cross section of the earth and thus,

Area = πr² where r is the radius of the earth given as: 6.371 x 10^(6)

A = π(6.371 x 10^(6)) ²

So I = 11.28 x 10^(7) x π(6.371 x 10^(6))² = 2.258 x 10^(17)A

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sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge q_{1}=9\times10^{-6}\ C at origin

Second charge q_{2}=6\times10^{-6}\ C

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

E=\dfrac{kq}{r^2}

0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}

\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}

\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1

\dfrac{1}{d}=1.82

d=\dfrac{1}{1.82}

d=0.55\ m

Hence, The coordinates of the point is (0,0.55).

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A runner starts from rest and in 3 s reaches a speed of 8 m/s. If we assume that the speed changed at a constant rate (constant
Stells [14]

Answer:

The average speed of the runner is 4 m/s.

Explanation:

Hi there!

The average speed (a.s) is calculated by dividing the traveled distance (d) over the time needed to travel that distance (t):

a.s = d / t

So, let´s find the distance traveled in those 3 s. For that, we can use the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at the point where the runner starts, then, x0 = 0. Since the runner starts from rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

We have the time (3 s), so let´s find the acceleration. For that, we can use the equation of velocity of an object moving in a straight line with constant acceleration:

v = v0 + a · t

Where "v" is the velocity at a time "t". Since v0 = 0, then:

v = a · t

At t = 3 s, v = 8 m/s

8 m/s = a · 3 s

8/3 m/s² = a

So let´s find the position of the runner at t = 3 s (In this case, the position of the runner will be equal to the traveled distance):

x = 1/2 · a · t²

x = 1/2 · 8/3 m/s² · (3 s)²

x = 12 m

Now, we can calcualte the average speed:

a.s = d/t

a.s = 12 m / 3 s

a.s = 4 m/s

The average speed of the runner is 4 m/s.

4 0
3 years ago
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