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3 years ago

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Consider a spherical Gaussian surface and three charges: q1 = 1.60 μC , q2 = -2.61 μC , and q3 = 3.67 μC . Find the electric flu

x through the Gaussian surface if it completely encloses (a) only charges q1 and q2, (b) only charges q2 and q3, and (c) all three charges.

1 answer:

Olenka [21]3 years ago

4 0

Answer:

Explanation:

<h3>Guass Law: Also known as "Gauss's flux theorem" is the total of the electric flux "φ" out of a closed surface is equal to the charge "Q" enclosed divided by the permittivity εο. Solution is attached.</h3>

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Answer:

1) Q ’= 8 Q , 2) q ’= 16 q , 3) r ’= ¾ r

Explanation:

For this exercise we will use Coulomb's law

F = k q Q / r²

It asks us to calculate the change of any of the parameters so that the force is always F

Original values

q, Q, r

Scenario 1

q ’= 2q

r ’= 4r

F = k q ’Q’ / r’²

we substitute

F = k 2q Q ’/ (4r)²

F = k 2q Q '/ 16r²

we substitute the value of F

k q Q / r² = k q Q '/ 8r²

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Q ’= Q

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Scenario 3

q ’= 3/2 q

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we substitute

k q Q r² = k (3/2 q) (⅜ Q) / r’²

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3 years ago

If the number of particles is increased in a balloon what happens to the pressure inside of it?

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The air pressure inside the balloon increases as the number of particles increases.

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3 years ago

A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

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There is a missing data in the text of the problem (found on internet):
"with force constant<span> k=</span>450N/<span>m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
</span> $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
<span>where
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x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
</span> $E=K_{max} = \frac{1}{2}mv_{max}^2$
<span>Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
</span> $E=U_{max}= \frac{1}{2}k A^2$
<span>
Since the mechanical energy must be conserved, we can write
</span> $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
<span>from which we find the maximum speed
</span> $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$
<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
</span> $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span> $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
<span>And since
</span> $E=U+K$
<span>we find the kinetic energy when the glider is at this position:
</span> $K=E-U=0.36 J - 0.05 J = 0.31 J$
<span>And then we can find the corresponding velocity:
</span> $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$
<span>
c) </span><span>the magnitude of the maximum acceleration of the glider;
</span>
For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m

For a simple harmonic motion, the acceleration is given by
</span> $a(t)=\omega^2 x(t)$
<span>where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
</span> $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$
<span>
e) </span><span>the total mechanical energy of the glider at any point in its motion. </span><span>

we have already calculated it at point b), and it is given by
</span> $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

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Determine a formula for the acceleration of the system in terms of mA, mB, θ, and g. Ignore the mass of the cord and pulley. Exp

jekas [21]

Answer:

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Explanation:

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Explanation:

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