Answer:
L = 130 decibels
Explanation:
The computation of the sound intensity level in decibels is shown below:
According to the question, data provided is as follows
I = sound intensity = 10 W/m^2
I0 = reference level = 
Now
Intensity level ( or Loudness)is




Therefore
L = 13 bel
And as we know that
1 bel = 10 decibels
So,
The Sound intensity level is
L = 130 decibels
Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
Answer:
141.56 N.
Explanation:
Data given:
Weight of the box= 200.2 N
Angle with the horizontal= 37.1°
Solution;
Gravitational force on the box,
= weight of the box
= 200.2 N
Component of gravitational force along plane =
( ∅ )
= W * (sin∅)
= (200.1) * sin (37.1°)
= 141.56 N
Answer:

and

Explanation:
Given:
- first charge,

- second charge,

- position of first charge,

- position of second charge,

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.
<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

- since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.





Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

and

Answer:
25.33 rpm
Explanation:
M = 100 kg
m1 = 22 kg
m2 = 28 kg
m3 = 33 kg
r = 1.60 m
f = 20 rpm
Let the new angular speed in rpm is f'.
According to the law of conservation of angular momentum, when no external torque is applied, then the angular momentum of the system remains constant.
Initial angular momentum = final angular momentum
(1/2 x M x r^2 + m1 x r^2 + m2 x r^2 + m3 x r^2) x ω =
(1/2 x M x r^2 + m1 x r^2 + m3 x r^2 ) x ω'
(1/2 M + m1 + m2 + m3) x 2 x π x f = (1/2 M + m1 + m3) x 2 x π x f'
( 1/2 x 100 + 22 + 28 + 33) x 20 = (1/2 x 100 + 22 + 33) x f'
2660 = 105 x f'
f' = 25.33 rpm