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Sergio [31]
3 years ago
6

Consider the space between a point charge and the surface of a neutral spherical conducting shell. If the charge sits at the cen

ter of the spherical shell, then the electric field between the two, as well as the field outside the outer boundary of the conductor, is the same as the field you would measure if the conducting shell was not there, though the charges of the conductor will redistribute themselves to ensure zero E field inside the conductor.
True or False?
Physics
1 answer:
mr Goodwill [35]3 years ago
8 0

Answer: True

Explanation: A neutral spherical conducting shell, has no net electric field inside it. The neutral conductor separates its positive and negative charges, when it is kept in a region of electric field, so that the net electric field inside the conductor becomes zero

Let us assume that a spherical Gaussian sphere surrounding the cavity and inside the conductor.

Since electric field inside the conductor doesn't exists, therefore the net electric flux through the Gaussian surface is zero.

From Gauss's law, when net electric flux through the closed surface is zero, the net enclosed charge should be zero

In order to make net enclosed charge as zero inside the metal, the interior surface of the conductor acquires a charge of -q

Since the interior surface of the conductor acquired -q charge, in order to maintain the electrical neutrality of the conductor, the exterior surface of the conductor acquires +q charge on it

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Which of Newton's laws explains why satellites need very little fuel to stay in oribit?
Angelina_Jolie [31]

Sattelites don't need any fuel to stay in orbit. The applicable law is...."objects in motion tend to stay in motion". Having reached orbital velocity, any such object is essentially "falling" around the earth. Since there is no (or at least very little) friction in the vacuum of space, the object does not slow.... It simply continues.


Sattelites in "low" earth orbit do encounter some friction from the very thin upper atmosphere, and they will eventually "decay".

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4 0
3 years ago
A speed-time graph is shown below:
Juliette [100K]

Answer:

It traveled 4 centimeters.

Explanation:

In a speed versus time graph, the distance travelled is given by the area under the graph.

In this graph we have the following:

- The speed of the object is v = 1 cm/s between time t = 0 s and t = 4 s

- The speed of the object is v = 0 cm/s between time t = 4 s and t = 8 s

Since the speed in the second part is zero, the distance travelled in the second part is zero. So, the only distance travelled by the object is the distance travelled during the first part, which is equal to the area of the first rectangle:

d=v\Delta t=(1)(4-0)=4 cm

4 0
3 years ago
If a marble is released from a height of 10 meters how long would it take to hit the ground?
zlopas [31]
S = ut + 0.5at^2 

<span>10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414}  </span>

<span>t^2 ~ 2.04 </span>

<span>t ~ 1.43 seconds</span>
3 0
3 years ago
The 0.5kg soccer ball moves toward the net with a force of 4N. What is its acceleration?​
Eduardwww [97]

Answer:

8 m/s²

Explanation:

Given,

Force ( F ) = 4 N

Mass ( m ) = 0.5 kg

To find : -

Acceleration ( a ) = ?

Formula : -

F = ma

a = F / m

= 4 / 0.5

= 40 / 5

a = 8 m/s²

It's acceleration is 8 m/s².

6 0
3 years ago
Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
jenyasd209 [6]

The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.

2. This is immediate from (2.2.7).

3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =

3 0
3 years ago
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