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Zielflug [23.3K]
3 years ago
5

What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radiu

s r1 = 0.775 m ?
Physics
1 answer:
alexira [117]3 years ago
7 0
The expression for the flux through a sphere is
flux = charge/Enot (i.e. epsillon not: a constant)

this shows that flux doesn't vary with radius but with the charge of the points
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During an auto accident, the vehicle’s air bags deploy and slow down the passengers more gently than if they had hit the windshi
const2013 [10]

Answer:

0.381 m

Explanation:

Distance traveled S is found by

S = ut + \frac{1}{2}a{t^2}

Where S is distance traveled, u is initial velocity, t is time, a is acceleration

Since we acceleration a is given as 60g, where g is gravitational constant of 9.81 then a=60*9.81=588.6

The initial velocity u is zero hence ut=0

Substituting a with 588.6, t with 36 ms

\begin{array}{c}\\S = 0 + \frac{1}{2}\left( {588.6 {\rm{m/}}{{\rm{s}}^2}} \right){\left( {\left( {36 {\rm{ms}}} \right)\left( {\frac{{1 {\rm{s}}}}{{{{10}^3} {\rm{ms}}}}} \right)} \right)^2}\\\\ = 0.3814128
{\rm{m}}\\\end{array}  

S=0.381 m

4 0
3 years ago
A 540 kg satellite moves through deep space with a speed of 27 m/s. A booster rocket on the satellite fires for 1.4 s, giving a
prohojiy [21]

Answer: a

Explanation: because the answer is 1.4444444 and that's the closest

6 0
3 years ago
Read 2 more answers
Write an equation for the nth term of the arithmetic sequence: 64, 78, 92, 106, ...
Anna [14]

If  64  is term #1 of the sequence,
then the general term is
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4 0
4 years ago
(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

The charge chould be moved to infinity

7 0
4 years ago
An elevator has a mass of 3000 kg. If the upward tension in the
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Answer: 18 m

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