We can define power as the rate of doing work, it is the work done in unit time. The SI unit of power is Watt (W) which is joules per second (J/s). Sometimes the power of motor vehicles and other machines are given in terms of Horsepower (hp) which is approximately equal to 745.7 watts.
Power is the rate at which a force is applied to an object for example.current wire
Answer:
D) This is the correct answer
Explanation:
In this exercise the two ball loads are suspended by a thread.
To answer this exercise, let us remember that charges of the same sign repel and charges of a different sign attract.
Therefore, for the system to maintain equilibrium, the two charges must be of the same sign.
When examining the different proposals
A) in this case, as a sphere has no charge, there is no electric force and the induced charge is of the opposite sign, so the spheres attract each other
B) in this case there is an electric force, but being of a different sign, the force is attractive so the system is not in equilibrium
C) as the charges are of different magnitude the system does not have equal angles
D) This is the correct answer, since the charges have the same magnitude and are of the same sign, so the force is repulsive and is counteracted by the weight component
F_e = W sin θ
Answer:
This is because the 11 positive protons and 10 negative electrons end up with an overall charge of +1.
Explanation:
Answer:
The value of A is 1.5m/s^2 and B is 0.5m/s^³
Explanation:
The mass of the rocket = 2540 kg.
Given velocity, v(t)=At + Bt^2
Given t =0
a= 1.50 m/s^2
Now, velocity V(t) = A*t + B*t²
If, V(0) = 0, V(1) = 2
a(t) = dV/dt = A+2B × t
a(0) = 1.5m/s^²
1.5m/s^² = A + 2B × 0
A = 1.5m/s^2
now,
V(1) = 2 = A× 1 + B× 1^²
1.5× 1 +B× 1 = 2m/s
B = 2-1.5
B = 0.5m/s^³
Now Check V(t) = A× t + B × t^²
So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² × 1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s
Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)
