Max preassure = force / min area
= 3N / 0.1 x 0.05
= 600N/m(squared)
Copy off of the picture below itll help better, its what someone sent me when i asked this question
Answer:
1. e m fmax = 0.00598 Volt
2. Imax = 0.000854 Amp
Explanation:
1. Find the maximum induced emf.
e m fmax =
Given that e m fmax = N*A*B*w
N = 1
A = 2 cm^2 = 0.0002 m^2
f = 143 rotation per minute = 143/min
f = (143/min) * (1 min/60 sec) = 2.38/sec
w = 2Πf = 2 * Π * 2.38 = 14.95 rad/sec
B = 2T
e m fmax = N*A*B*w
e m fmax = 1 * 0.0002 * 2 * 14.95
e m fmax = 0.00598 Volt.
2. Find the maximum current through the bulb.
Imax = e m fmax / R
Where R is the total resistance in the circuit is 7 Ω.
Imax = 0.00598/7 = 0.000854 Amp.
Imax = 0.000854 Amp
(Missing figure is here: https://www.physicsforums.com/attachments/ch05-p070-jpg.149243/ )
Let's call

and

the masses of the two blocks. We can write Newton's second law for both blocks (sum of all forces acting on the block = ma). On block 1, we have two forces: the weight

pointing downwards and the tension of the string T poiting upwards. On block 2, we have the tension of the string going right and the friction

going left. Therefore


Summing the two equations, we find

and then using

we can find the acceleration:
Answer:
maximum amplitude = 0.08 m
Explanation:
Given that
Time period T= 0.58 s
acceleration of gravity g= 9.8 m/s²
We know that time period of simple harmonic motion given as
T = 2π/ω
0.58 = 2π/ω
ω = 10.83rad/s
ω=angular frequency
Lets take amplitude = A
The maximum acceleration given as
a= ω² A
The maximum acceleration should be equal to g ,then block does not separate
a= ω² A
9.8 = 10.83² A
A = 0.08m
maximum amplitude = 0.08 m