A systematic error in data is called

Physics

1 answer:

d1i1m1o1n [39]4 years ago

7 0

<span>an error having a nonzero mean, so that its effect is not reduced when observations are averaged.</span>

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Find the x and y coordinates of the position at which an electron would be in equilibrium.

love history [14]

Let e be the charge of the electron.
Then F1 + F2 = 0
F1 = K (-1.5) * (-e) /r1^2
F2 = K (15) * (-e) /r2^2
(-1.5) * /r1^2 + (15) /r2^2 = 0
-1 /r1^2 + (10) /r2^2 = 0
[r2/r1]^2 = 10
[R2/r1] = ± 3.162
r2 = ± 3.162 r1.

Given r1 + r2 = √ [1^2 + 0.5^2] = 1.1180-----1
Also tan θ = 0.5 => θ = 26.57°
====================================
Substituting r2 = 3.162 r1 in 1
4.162 r1=1.1180
r1 = 0.2686 m

x coordinate = r1 cos θ = 0.2403 m
y coordinate = r1 sin θ = 0.1201m
===================================

Substituting r2 = - 3.162 r1 in 1
-2.162 r1=1.1180
r1 = -0.5171m
x coordinate = r1 cos θ = -0.4625m
y coordinate = r1 sin θ = -0.2313m
===============================

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4 years ago

The FIRST step in cloud formation is.... A) warmed air cools and condenses. B) solar energy causes warm air to rise. C) solar en

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c because the fist step is condensing then rising up

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3 years ago

The elastic extensibility of a piece of string is .08. If the string is 100 cm long, how long will the string be when it is stre

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Answer:

The elastic extensibility of a piece of string is .08. If the string is 100 cm long, how long will the string be when it is stretched to the point where it becomes plastic? is your ansewer dont take tension

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A thick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.46 m, and is initially unc

KonstantinChe [14]

(a) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can solve the different part of the problem by using Gauss theorem.

Considering a Gaussian spherical surface with radius r<a (inside the shell), we can write:

$E(r) \cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

where q is the charge contained in the spherical surface, so

$q=5.00 C$

Solving for E(r), we find the expression of the field for r<a:

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

(b) 0

The electric field strength in the region a < r < b is zero. This is due to the fact that the charge +q placed at the center of the shell induces an opposite charge -q on the inner surface of the shell (r=a), while the outer surface of the shell (r=b) will acquire a net charge of +q.

So, if we use Gauss theorem for the region a < r < b, we get

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$

however, the charge q' contained in the Gaussian sphere of radius r is now the sum of the charge at the centre (+q) and the charge induced on the inner surface of the shell (-q), so

q' = + q - q = 0

And so we find

E(r) = 0

(c) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

We can use again Gauss theorem:

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$ (1)

where this time r > b (outside the shell), so the gaussian surface this time contained:

- the charge +q at the centre

- the inner surface, with a charge of -q

- the outer surface, with a charge of +q

So the net charge is

q' = +q -q +q = +q