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sasho [114]
3 years ago
11

Which area on the line has both some kinetic energy and some gravitational energy?

Physics
2 answers:
I am Lyosha [343]3 years ago
8 0
A or b be I believe
Yakvenalex [24]3 years ago
4 0
I believe it is either A or B
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Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance
siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

       

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

          E_ {total} = E₁ + E₂

          E_{ total} = k \frac{Q}{x_1^2} + k  \frac{Q}{x_2^2}

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = k  \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )

            E_ {total} = k \frac{Q}{x_1^2}   ( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

             x₁² =  (x₁ + d)²

           

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

              E_ {total} = -k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}

this expression holds for the points located at

                  -r₁ <x₁ <r₁

3) Point between the two spheres

                E_ {total} = k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

                  E_ {total} = + k \frac{Q}{(d-x_1)^2}+ k Q / (d-x1) 2

point range

                  -r₂ <x₂ <r₂

             

5) Right side point

            E_ {total} = k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}

             E_ {total} = - k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )- k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0
2 years ago
A sound has 13 crests and 15 troughs in 3 seconds. When the second crest is produced the first is 2cm away from the source? Calc
Yuki888 [10]

Answer:

The wavelength will be 4 cm, frequency will be 4.66 Hz and wave speed is 18.6 cm/sec

Explanation:

Given:

No. of crest = 13

No. of trough = 15

Time = 3 seconds

Hence, 1 crest or 1 trough = \frac{1}{2}\lambda

therefore,

13 C + 15 T = 28(\frac{1}{2}\lambda)

=14\lambda

Time given 3 seconds

  = \frac{3}{14}s

\nu= \frac{14}{3}

\nu= 4.6 Hz \approx 5 Hz

2 cm distance is travelled is time period

\lambda = 4 cm

Again wave will travel in 1 T = 4 cm

wave speed v =\lambda\times\nu

= 4\times\frac{14}{3}

= 18.6 cm/s

5 0
3 years ago
V=I/R correctly expresses the relationship between voltage, current, and resistance.
docker41 [41]
The correct answer is true
5 0
1 year ago
Read 2 more answers
A Heavy crate applied a force of 1500 N on a 25m2 piston. What force needs to be applied on a 0.8m2 piston to lift the crate?
Aleks04 [339]
25/1500 is equal to 0.8/x
0.8*1500 is equal to 1200
1200/25 is equal to 48 N
7 0
3 years ago
Read 2 more answers
The first PCS technology used a form of time division multiplexing called ____.​ a. ​Time Division Multiple Access (TDMA) b. ​Gl
salantis [7]

Answer:

A. Time Division multiple Access (TDMA)

Explanation:

Time-division multiple access (TDMA) is a channel access method for shared-medium networks

It allows several users to share the same frequency channel by dividing the signal into different time slots. The users transmit in rapid succession, one after the other, each using its own time slot.

3 0
3 years ago
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