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Velocity
Note : Not speed as Acceleration is a vector!
Hey user
The energy E in joules (J) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C):
E(J) = V(V) ×<span> Q</span>(C)
So
joule = volt × coulomb
or
J = V × C
Example
What is the energy in joules that is consumed in an electrical circuit with voltage supply of 15V and charge flow of 4 coulombs?
E = 15V × 4C = 60J
Answer:
The displacement of the car after 6s is 43.2 m
Explanation:
Given;
velocity of the car, v = 12 m/s
acceleration of the car, a = -1.6 m/s² (backward acceleration)
time of motion, t = 6 s
The displacement of the car after 6s is given by the following kinematic equation;
d = ut + ¹/₂at²
d = (12 x 6) + ¹/₂(-1.6)(6)²
d = 72 - 28.8
d = 43.2 m
Therefore, the displacement of the car after 6s is 43.2 m
Answer:
b) 4781 N
Explanation:
Because there is a redius do this question is talking about the acceleration force which= mv^2/r
so a=15^2/80=2.8125 m^2/s
so the force will be = m.a
F =1700×2.8125=4781.25 N
Answer:
1-In a uniform electric field, the field lines are straight, parallel, and uniformly spaced this statement is true.
2-Electric field lines near positive point charges radiate outward. this statement is also true.
3-The electric force acting on a point charge is proportional to the magnitude of the point charge. this statement is true as well.
Explanation:
the electric field created by a point charge is defined by E=KQ/r^2 where k is constant, q is magnitude of charge and r is the distance away from the point charge so the electric filed is distance dependent and can not be constant at all distances.
electric field lines near a negative point charge are directed radially inward because negative charge attracts the field and is not clockwise.
