Hi there!
Recall the equation for electric potential of a point charge:
V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)
Q = Charge (C)
r = distance (m)
We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.
Upper right charge's potential:
Lower left charge's potential:
Add the two, and subtract from the total EP at the point:
The remaining charge must have a potential of 2036.25 V, so:
Answer:
Explanation:
Parameters given:
Charge of object, q = 5 mC = 
Acceleration of object, a = 
Mass of object, m = 2.0 g
The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).
We know that Electrostatic force, F, is given in terms of Electric field, E, as:
F = qE
This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).
The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:
F = ma
Equating the two forces of the object, we get:
-qE = ma
=> 
Solving for E, we have:
The magnitude will be:
The electric field has a magnitude of 0.002 N/C.
To find work, you use the equation: W = Force X Distance X Cos (0 degrees)
Following the Law of Conservation of Energy, energy cannot be destroyed nor created.
So you would do 75 N x 10m x Cos (0 degrees)= 750 J
Answer:
A. reintroducing an animal to the ecosystem
Explanation:
As generally, all know that for restoring an ecosystem naturally, it requires reintroduction of an animal to the ecosystem. As though it helps in reimposing the ecosystem back, and also helps to improve our ecosystem in natural surroundings, natural terrain, and population density. Basically reintroducing an animal is also required for the balancing of the ecosystem. As everything requires a properly balanced nature.
Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is
<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²
where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².
The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is
<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²
<em>F</em> ≈ 9.81 N
Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.
This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of
<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²
i.e. 1/4 of the weight on Earth, which would be about 2.45 N.
