Hello!
Using the equation for the electric field produced by a source charge:
E = Electric Field Strength ( 2.86 × 10⁵ N/C)
k = Coulomb's Constant ( 8.99 × 10⁹ Nm²/C²)
q = Charge of source charge (3 μC = 0.000003 C)
r = distance of test charge from source charge (m²)
We can rearrange the equation to solve for distance to make plugging in values easier. (Isolate for 'r').
Plug in the given values.
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Answer:
The speed of the car at the end of the 2nd second = 8.0 m/s
Explanation:
The equations of motion will be used to solve this problem.
A car starts from rest,
u = initial velocity of the car = 0 m/s
Accelerates at a constant rate in a straight line,
a = constant acceleration of the car = ?
In the first second the car moves a distance of 2.0 meters,
t = 1.0 s
x = distance covered = 2.0 m
x = ut + (1/2)at²
2 = 0 + (1/2)(a)(1²)
a = 4.0 m/s²
How fast will the car be moving at the end of the second second
Now,
a = 4.0 m/s²
u = initial velocity of the car at 0 seconds = 0 m/s
v = final velocity of the car at the end of the 2nd second = ?
t = 2.0 s
v = u + at
v = 0 + (4×2)
v = 8.0 m/s
Answer:
Q = 0.24
Explanation:
given data:
resonant angular frequency is given as \omega_O = \frac{1}{\sqrt{LC}}
where L is inductor = 150 mH
C is capacitor = 0.25\mu F
QUALITY FACTOR is given as
Putting all value to get quality factor value
Q =
Q = 0.24