I believe this is known as wave period.
hope this helps!
Answer:0.6kw
Explanation:
Power=force×velocity
Power=20×30=600w
In kw it's going to be 600/1000=0.6kw
Answer:
E=
Explanation:
We are given that
Charge on ring= Q
Radius of ring=a
We have to find the magnitude of electric filed on the axis at distance a from the ring's center.
We know that the electric field at distance x from the center of ring of radius R is given by

Substitute x=a and R=a
Then, we get




Where K=
Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=
Answer: the minimal force that you need to apply to move the bureau is F = 198.45N
Explanation:
If you want to move an object, you need to apply a force that is bigger than the force of the statical friction.
The force of statical friction can be written as.
Ff = k*N
where k is the coefficient of static friction, in this case, k = 0.45, and N is the normal force between the object and the surface.
In this case, the normal force is the weight of the bedroom bureau, this is:
N = m*g = 45kg*9.8m/s^2 = 441N
Then the force is:
Fr = 0.45*441N = 198.45N
This means that the minimal force that you need to apply to move the bureau is F = 198.45N
and after this point, the force of friction will work wit the kinetic coefficient of friction, that usually is smaller than the statical one.
Answer:
0.4 m/s2
Explanation:
mass: 25kg
net force: 10 N
acceleration: ?
net force ÷ by mass= acceleration
10 N ÷ 25 Kg = 0.4 m/s2