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iren2701 [21]
3 years ago
12

How do electric field lines indicate the strength of the field?

Physics
2 answers:
Paha777 [63]3 years ago
6 0
I am almost sure it it (c)
KiRa [710]3 years ago
3 0
<h2>Answer:</h2>

The correct answer is the option B which is by how close the lines are.

<h3>Explanation:</h3>
  • Electric field of lines are a type of pattern used to describe the direction and the strength of the electric field.
  • These are lines which always projects from Positive ions to negative ions.
  • Then strength of electric field is shown by the closeness of lines.
  • Where these lines are close to each other it means strength of electric field is high and vice versa.
  • These lines show that electric field is a vector property.


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According to Newton's law of motion, when we shake a mango tree, mangoes fall down explain.​
Reptile [31]
When we shake a mango tree, the mangoes fall down. It is because when we shake the tree, the mango tend to be rest due to inertia where as the branches are in motion. That is why the mangoes tend to be at rest due to inertia where as the branches are in the motion.
8 0
3 years ago
For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant
myrzilka [38]

First let us imagine the projectile launched at initial velocity V and at angle θ relative to the horizontal. (ignore wind resistance)

Vertical component y:

The initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum height of h, the vertical velocity will be 0, therefore the time t taken to attain this maximum height is:

h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g

where g is  acceleration due to gravity

Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g] 
D = (2V²sinθ cosθ)/g 
 D = (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, dD/dθ = 0
That is,

2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2  or  θ = π/4


Therefore it is now<span> shown that the maximum horizontal travelled is attained when the launch angle is π/4 radians, or 45°.</span>

6 0
4 years ago
Solve the following system of equations by using either substitution or elimination.
strojnjashka [21]
I believe that the answer to the question provided above are the following;

x = 29.8410

y = 16.6794

z = -1.2642
Hope my answer would be a great help for you.    If you have more questions feel free to ask here at Brainly.
5 0
3 years ago
The membrane that surrounds a certain type of living cell has a surface area of 4.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
Nadusha1986 [10]

Answer:

The charge resides on the outer surface = 1.245 \times 10^{-12} C

Explanation:

Surface area of cell  (A) = 4.3\times 10^{-9}  m^{2}

Separation between two plate  (d) = 1.1 \times 10^{-8}  m  

Dielectric constant (k) = 4.2

Potential difference (\Delta V) = 85.7 \times 10^{-3} V

The capacitance of parallel plate capacitor in free space is given by,

           C = \frac{\epsilon_{o} A }{d}

Where \epsilon_{o}  = permittivity of free space = 8.85 \times 10^{-12}

The Capacitance of capacitor is increase by k times when it placed in dielectric medium.

C_{dielectric}  = \frac{k \epsilon_{o} A }{d}

And we know that, C = \frac{Q}{ \Delta V}

So charge on the outer surface is given by,

      Q = \frac{k \epsilon A \Delta V }{d}

      Q = \frac{4.2 \times 4.3 \times 10^{-9} \times 8.85 \times 10^{-12} \times 85.7\times 10^{-3}   }{1.1 \times 10^{-8} }

      Q = 1.245 \times 10^{-12}

3 0
4 years ago
An atom that has fewer neutrons than protons and more electrons than protons is a ion
riadik2000 [5.3K]

i think it is an ion

6 0
3 years ago
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