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BaLLatris [955]
3 years ago
9

The total electric flux from a cubical box 26.0 cm on a side is 1840 N m2/C. What charge is enclosed by the box?

Physics
1 answer:
vitfil [10]3 years ago
7 0

The expression of the electric flux is

\Phi = \frac{Q}{\epsilon_0}

Here,

Q = Total charge enclosed in the closed surface

\epsilon_0 = Permittivity due to free space

Rearranging to find the charge,

Q = \epsilon_0 \Phi

Replacing with our values we have finally

Q = (8.85*10^{-12}F\cdot m^{-1})(1.84*10^3 N\cdot m^2/C)

Q = 1.6284*10^{-8} C (\frac{10^9nC}{1C})

Q = 0.1684nC

The charge enclosed by the box is 0.1684nC

The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.

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Answer:

v_0=9.9\ m.s^{-1}

Explanation:

Given:

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  • range of projectile, R=5\ m

<u>We have formula  for the range of projectile:</u>

R=\frac{v_0^2\times sin\ 2\theta}{g}

putting the respective values

5=\frac{v_0^2\times sin\ 30^{\circ}}{9.8}

v_0=9.9\ m.s^{-1} is the velocity with which Tom should jump to land on the other roof.

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3 years ago
How is the wavelength of a longitudinal wave determined?
astraxan [27]

Answer:

Wavelength can always be found by measuring the distance between any two corresponding points on adjacent waves. In the case of a longitudinal wave, a wavelength measurement is made by measuring the distance from a compression to the next compression or from a rarefaction to the next rarefaction.

Explanation:

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3 years ago
In the shadow of a tree with a dense, leafy canopy, one sees a number of light spots. Surprisingly, they all appear to be circul
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The characteristics of the diffraction phenomenon allow to find the result for the shape of the points of light that you pass the tree is:

  • The shape of the dots is circular because it is in the range of far-field diffraction.

Diffraction is the phenomenon where the undulatory part of the light becomes evident, it is the interference of the waves that make up each ray of light, for this phenomenon to occur it must be fulfilled that the wavelength is of the order of the space where pass the light.

In the leafy tree it has many leaves, but there are spaces between them, some of these spaces are small and it fulfills the diffraction condition, therefore we see bright spots and not a continuous shadow.

Diffraction can be classified depending on the distance to the observer:

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In this case, the distance from the leaves to the observer is large, therefore we are in the case of far-field diffraction and since the edge of the leaves that forms the diffraction is closed, the observable shape is a circle.

In conclusion using the characteristics of the diffraction phenomenon we can find the result for the shape of the points of light that pass the tree is:

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Learn more about diffraction here:  brainly.com/question/20140459

8 0
2 years ago
When a light wave enters into a medium of different optical density,
olga nikolaevna [1]
I think it should be option (b)
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3 years ago
An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric
8090 [49]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

3 0
3 years ago
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