Answer:
The maximum voltage is 41.92 V.
Explanation:
Given that,
Peak voltage = 590 volts
Suppose in an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is
.
We need to calculate the resonance frequency
Using formula of frequency

Put the value into the formula


We need to calculate the maximum current
Using formula of current




Impedance of the circuit is

At resonance frequency 

We need to calculate the maximum voltage
Using ohm's law



Hence, The maximum voltage is 41.92 V.
Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
Answer:
3ohms
Explanation:
From Ohm's Law
V = IR
V is that voltage = 3volts
I = current = 1amp
R = resistance in ohms
Putting those values into the above formula.
3volts = 1amp×R
Making R the subject
R = 3/1
R = 3ohms
The resistance of the light bulb is 3ohms.
Answer:
128.9 N
Explanation:
The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

where
F is the force
is the change in momentum
is the time interval
The change in momentum can be written as

where
m = 0.04593 kg is the mass of the ball
u = 0 is the initial velocity of the ball
is the final velocity of the ball
Substituting into the original equation, we find the force exerted on the golf ball:

<span> the </span>electric field<span> direction about a </span>positive<span> source </span>charge<span> is always directed away from the </span>positive<span> source. And the </span>electric field <span>direction about a negative source </span>charge<span> is always directed toward the negative source.</span>