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2 years ago

Gauss's law is usualy written as :

Physics

1 answer:

snow_tiger [21]2 years ago

4 0

Answer:

(a) the net charge inside the closed surface.

Explanation:

In Gauss' Law, Qencl refers to the net charge inside the Gaussian surface. This surface is usually taken as a symmetric geometric surface, but this is merely for simplicity. Gauss' Law holds for any closed surface. Inside this surface there can be insulators as well as conductors. Regardless of the geometry or the materials inside, Qencl refers to the net charge inside the closed surface. The charge outside the surface is irrelevant for Gauss' Law, therefore all the charge in the physical system is not included in Gauss' Law.

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An angry physics student releases a wrecking ball as shown. The wrecking ball is just about to hit the building at the final tim

daser333 [38]

Answer:

the force between the building and the ball is non-conservative (friction-type force)

Explanation

Explanation:For this exercise the student must create an impulse to move the ball towards the building, in this part he performs positive work since the applied force and the displacement are in the same direction.

When the ball moves it has a kinetic energy and if its height increases or decreases its potential energy also changes, but the sum of being must be equal to the initial work.

When the ball arrives and collides with the building, non-conservative forces, of various kinds; rubbing, breaking, etc. It transforms this energy into a part of heat and another in mechanical energy that the building must absorb, let us destroy its wall

Consequently, the force between the building and the ball is non-conservative (friction-type force

6 0

2 years ago

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

3 years ago

Is the change in position of an object

natima [27]

Displacement is the change in position

6 0

3 years ago

Read 2 more answers

1 hp is equal to? What is the answer

rewona [7]

The electrical equivalent of one horsepower is 746 watts in the International System of Units (SI), and the heat equivalent is 2,545 BTU (British Thermal Units) per hour. Another unit of power is the metric horsepower, which equals 4,500 kilogram-metres per minute (32,549 foot-pounds per minute), or 0.9863 horsepower.

5 0

2 years ago

Is v2 = v1t+a dimensionally correct? Explain please!

Lady bird [3.3K]

You want v2 = v1 + at
v is measured in m/s, a in m/s2, and t in s.
the dimensions multiply like algebraic quantities.
so because v2 is measured in m/s, then (v1 + at) has to come out in m/s
the units for (v1 + at) are (m/s) + (m/s2)(s)
time "s" cancels out one acceleration "s", so it comes ut to (m/s) + (m/s), which = (m/s).
if you had (v1t + a), then you would have (m/s)(s) + (m/s2) which = (m) + (m/s2), which doesn't work.

4 0

3 years ago