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Juliette [100K]
3 years ago
13

A cube with sides of area 22 cm2 contains a 21.2 nanoCoulomb charge. Find the flux of the electric field through the surface of

the cube in unis of Nm2/C. Enter a number with one digit behind the decimal point.
Physics
1 answer:
VladimirAG [237]3 years ago
3 0

Answer:

Explanation:

According to Gausses theorem, electric flux coming out of  charge   on all sides q is equal to

q / ε₀

q is charge , ε₀  is permittivity of air whose value is given by

8.85 x 10⁻¹²; q = 21.2 x 10⁻⁹ C

Putting the values we get

FLUX = \frac{21.2\times10^{-9}}{8.85\times10^{-12}}

2.4 x 10³ Nm²/C

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A pendulum is made by letting a 4 kg mass swing at the end of a string that has a length of 1.5 meter. The maximum angle that th
olga nikolaevna [1]

Answer:

Approximately 7.8\; \rm J.

Explanation:

The change in the gravitational potential energy of the pendulum is directly related to the change in its height.

Refer to the sketch attached. The pendulum is initially at \rm P_2. Its highest point is at P_1. The length of segment \rm BP_2 gives the change in its height.

The lengths of \rm AP_1 and \rm AP_2 are simply the length of the string, 1.5\; \rm m. To find the length of \rm BP_2, start by calculating the length of \rm AB.

\rm AB forms a leg in the right triangle \rm \triangle AP_1B. Besides, it is adjacent to the 30^\circ angle \rm P_1\hat{A}B. Its length would be:

\rm AB = 1.5 \times \cos(30^\circ) \approx 1.30\; \rm m.

The length of \rm BP_2 would thus be

\rm BP_2 = AP_2 - AB = 1.5 - 1.30 \approx 0.20\; \rm m.

The change in gravitational potential energy can be found with the equation

\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h. In this equation,

  • m is the mass of the object,
  • g \approx 9.81\; \rm N \cdot kg^{-1} near the surface of the earth, and
  • \Delta h is the change in the object's height.

In this case, m = 4\; \rm kg and \Delta h \approx 0.20\; \rm m. Therefore:

\Delta \mathrm{GPE} = 4 \times 9.81 \times 0.20 \approx 7.8\; \rm J.

6 0
3 years ago
Consider a 100 g object dropped from a height of 1 m. Assuming no air friction (drag), when will the object hit the ground and a
Katyanochek1 [597]

Answer:

speed and time are Vf = 4.43 m/s and  t = 0.45 s

Explanation:

This is a problem of free fall, we have the equations of kinematics

      Vf² = Vo² + 2g x

As the object is released the initial velocity is zero, let's look at the final velocity with the equation

      Vf = √( 2 g X)

      Vf = √(2 9.8  1)

      Vf = 4.43 m/s

This is the speed with which it reaches the ground

 

Having the final speed we can find the time

      Vf = Vo + g t

       t = Vf / g

       t = 4.43 / 9.8

       t = 0.45 s

This is the time of fall of the body to touch the ground

3 0
3 years ago
A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of
Deffense [45]

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

  1.29 kg/m³   =  <u>      mass    </u>

                             1800 m³

mass =  1.29 kg/m³  ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

8 0
3 years ago
Read 2 more answers
Where are the oldest rocks found ?
nignag [31]

Answer:

zircons

Explanation:

they are over 4.375 billion years old

7 0
3 years ago
A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a
Stella [2.4K]

Answer:

y = 67.6 feet,   y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

         tan θ = y / L

For the small triangle. Projector up to the person

         tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

         y₀ / (L -d) = y / L

         y = y₀  L / (L-d)

The distance from the screen (d), we look for it with kinematics

         v = d / t

        d = v t

we replace

         y = y₀ L / (L - v t)

         y = 5.2 22 / (22 - 3 t)

         y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

           y = 114.4 / (22-13)

           y = 67.6 feet

7 0
3 years ago
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