RANGKAIAN RLC SERI PADA TEGANGAN BOLAK BALIK 220V R=80 OHM XL= 90 OHM XC = 30 OHM MAKA TEGANGAN PADDA INDUKTOR ADALAH???
1 answer:
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Answer:
Vi = 32 [m/s]
Explanation:
In order to solve this problem we must use the following the two following kinematics equations.
The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.
where:
Vf = final velocity = 8[m/s]
Vi = initial velocity [m/s]
a = acceleration = [m/s^2]
t = time = 5 [s]
Now replacing:
8 = Vi - 5*a
Vi = (8 + 5*a)
As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.
where:
d = distance = 100[m]
(8^2) = (8 + 5*a)^2 - (2*a*100)
64 = (64 + 80*a + 25*a^2) - 200*a
0 = 80*a - 200*a + 25*a^2
0 = - 120*a + 25*a^2
0 = 25*a(a - 4.8)
therefore:
a = 0 or a = 4.8 [m/s^2]
We choose the value of 4.8 as the acceleration value, since the zero value would not apply.
Returning to the first equation:
8 = Vi - (4.8*5)
Vi = 32 [m/s]
Answer:
Explanation:
F = Force =
m = Mass of proton =
t = Time taken =
Acceleration is given by
The velocity of the proton is