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4 years ago

Why do the planets in our solar system have a layered internal structure?

1 answer:

4 0

This is because a planet has unique, internal characteristics or composition that are settled at different depths based on their given density.

Taken, for example, our own planet Earth. The Earth's structure is well determined. It's interior consists of several layers: the crust- a relatively thin region of low-density silicates; the mantle- a thick region of higher-density of iron-rich silicates, and lastly is the core- the central region of iron which is mixed with various impurities.

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Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

4 years ago

Show that a directed multigraph having no isolated vertices has an euler circuit if and only if the graph is weakly connected an

sergeinik [125]

If we let
p as the directed multigraph that has no isolated vertices and has an Euler circuit
q as the graph that is weakly connected with the in-degree and out-degree of each vertex equal

The statement we have to prove is
p ←→q (for biconditional)
Since
p → q (assuming that p is strongly connected to q)
q ← p (since p is strongly connected to q)
Therefore, the bicondition is satisfied

5 0

3 years ago

What is the pooled variance for the following two samples?

Snowcat [4.5K]

I think it is B as 168/20

4 0

3 years ago

An echo repeats two syllables.If the velocity of sound is 330 m/s , then the distance of the reflecting surface is-

Andreas93 [3]

Well, the main point you should remember is that we only can hear 20hz that leads to T=0.05sec which means that the distance leads us to velocity x time.
When calculating, you get 330 * 0.05=16.5 .
The number which you got from the previous step has to me multiplied 4 times (4 minutes) and you will get the answer you need - a) 66.0 m.
Do hope it will help you! Regards!

6 0

4 years ago

Please help!!!!!!!!!!!!!!!

jenyasd209 [6]

7.1. The graph displays velocity over time, so the distance covered by "him" is equal to the unsigned (positive) area under the curve. (In contrast, the signed area represents displacement.) Finding this area is just an exercise in basic geometry.

• From time 0 to 3 s, the distance is equal to the area of a triangle with height 15 m/s and length 3 s:

1/2 (15 m/s) (3 s) = 22.5 m

• From 3 to 5.5 s, the distance is the area of a rectangle with height 15 m/s and length 5.5 s - 3 s = 2.5 s:

(15 m/s) (2.5 s) = 37.5 m

• From 5.5 to 6.5 s, you have a trapezoid with "bases" 15 m/s and 5 m/s, and "height" 6.5 s - 5.5 s = 1 s:

1/2 (15 m/s + 5 m/s) (1 s) = 10 m

• From 6.5 to 8 s, you have a triangle with height 5 m/s and length 8 s - 6.5 s = 1.5 s:

1/2 (5 m/s) (1.5 s) = 3.75 m

• From 8 to 9 s, another triangle with height 13 m/s and length 9 s - 8 s = 1 s:

1/2 (13 m/s) (1 s) = 6.5 m

• From 9 to 13 s, a rectangle with height 13 m/s and length 13 s - 9 s = 4 s:

(13 m/s) (4 s) = 52 m

• From 13 to 16.5 s, a triangle with height 13 m/s and length 16.5 s - 13 s = 3.5 s:

1/2 (13 m/s) (3.5 s) = 22.75 m

Add up the distances to get the total:

22.5 m + 37.5 m + 10 m + 3.75 m + 6.5 m + 52 m + 22.75 m = 155 m

7.2. The velocity is non-zero for any given time interval, so "he" is never at rest. (True, his velocity is 0 at 8 s, but only instantaneously.)

7.3. Given the plot of velocity, the acceleration is negative wherever the slope of the tangent line to the curve is negative. This happens in the interval from 5.5 to 9 s.

7.4. Similarly, positive acceleration corresponds to a positively-sloped tangent line. This happens from 0 to 3 s, and again from 13 to 16.5 s.

7.5. Where the velocity curve is horizontal, the accleration is zero, so you can ignore those intervals.

• From 0 to 3 s, the acceleration is

(15 m/s - 0 m/s)/(3 s - 0 s) = 5 m/s²

• From 5.5 to 6.5 s, it is

(5 m/s - 15 m/s)/(6.5 s - 5.5 s) = -10 m/s²

• From 6.5 to 8 s, it is

(0 m/s - 5 m/s)/(8 s - 6.5 s) ≈ -3.3 m/s²

• From 8 to 9 s, it is

(-13 m/s - 0 m/s)/(9 s - 8 s) = -13 m/s²

• From 13 to 16.5 s, it is

(0 m/s - (-13 m/s))/(16.5 s - 13 s) ≈ 3.7 m/s²

The clear winner is the interval from 8 to 9 s, where the acceleration has a magnitude of 13 m/s².

8. The magnitude of the velocity of the ball decreases until it reaches zero at its maximum height, then increases as it falls back down. Acceleration is constant and pointing downward the entire time.

4 0

3 years ago