Answer:
There are four main ways of doing that :-
- Velocity
- Acceleration
- Momentum
- Kinetic energy
Hope it helps!
Answer:
Not acted upon there is no motion
Forces acted upon there is motion
You can refer to the definition stated below:
Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia. ... If that velocity is zero, then the object remains at rest.
Answer:
2123.55 $/hr
Explanation:
Given parameters are:
KV
L = 143 km
I = 500 A
So, we will find the voltage potential provided for the city as:
kV
kV
Then, we will find dissipated power because of the resistive loss on the transmission line as:
W
Since the charge of plant is not given for electric energy, let's assume it randomly as 
Then, we will find the price of energy transmitted to the city as:
$/hr
To calculate money per hour saved by increasing the electric potential of the power plant:
Finally,
$/hr
The amount of money saved per hour = 
$/hr
Note: For different value of the price of energy, it just can be substituted in the equations above, and proper result can be found accordingly.
<h2>
Answer:</h2>
(a) 6.95 x 10⁻⁸ C
(b) 6.25N/C
<h2>
Explanation:</h2>
The electric field (E) on a point charge, Q, is given by;
E = k x Q / r² ---------------(i)
Where;
k = constant = 8.99 x 10⁹ N m²/C²
r = distance of the charge from a reference point.
Given from the question;
E = 10000N/C
r = 0.250m
Substitute these values into equation(i) as follows;
10000 = 8.99 x 10⁹ x Q / (0.25)²
10000 = 8.99 x 10⁹ x Q / (0.0625)
10000 = 143.84 x 10⁹ x Q
Solve for Q;
Q = 10000/(143.84 x 10⁹)
Q = 0.00695 x 10⁻⁵C
Q = 6.95 x 10⁻⁸ C
The magnitude of the charge is 6.95 x 10⁻⁸ C
(b) To get how large the field (E) will be at r = 10.0m, substitute these values including Q = 6.95 x 10⁻⁸ C into equation (i) as follows;
E = k x Q / r²
E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 10²
E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 100
E = 6.25N/C
Therefore, at 10.0m, the electric field will be just 6.25N/C
