Hooke's Law says that F=-kx where k is the spring constant measured in N/m (newtons per meter)
Answer:
The magnitude of F1 is
The magnitude of F2 is
And the direction of F2 is
Explanation:
<u>Net Force
</u>
Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.
The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.
Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:
The sum of the vertical components of F1 and F2 must equal the total force Ft
Solving for 
in the first equation
The magnitude of F1 is
The magnitude of F2 is
And the direction of F2 is
Answer:
James Chadwick
Explanation:
In May 1932 James Chadwick announced that the core also contained a new uncharged particle, which he called the neutron
(a) The time for the capacitor to loose half its charge is 2.2 ms.
(b) The time for the capacitor to loose half its energy is 1.59 ms.
<h3>
Time taken to loose half of its charge</h3>
q(t) = q₀e-^(t/RC)
q(t)/q₀ = e-^(t/RC)
0.5q₀/q₀ = e-^(t/RC)
0.5 = e-^(t/RC)
1/2 = e-^(t/RC)
t/RC = ln(2)
t = RC x ln(2)
t = (12 x 10⁻⁶ x 265) x ln(2)
t = 2.2 x 10⁻³ s
t = 2.2 ms
<h3>
Time taken to loose half of its stored energy</h3>
U(t) = Ue-^(t/RC)
U = ¹/₂Q²/C
(Ue-^(t/RC))²/2C = Q₀²/2Ce
e^(2t/RC) = e
2t/RC = 1
t = RC/2
t = (265 x 12 x 10⁻⁶)/2
t = 1.59 x 10⁻³ s
t = 1.59 ms
Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.
Learn more about energy stored in capacitor here: brainly.com/question/14811408
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