Answer:
The intensity of the light from the bulb would be
3.501 x 
W/
Explanation:
Given
The Power = 110 n W = 110 x 
W
the distance r = 50 mm = 50 /1000 = 0.05 m
The intensity can be obtained with the relationship below;
I = Power/area ......1
The area of the sphere would be used in this case since the bulb is spherical; A=4π
Putting it into equation 1, we have;
I = P/ 4π
I = 110 x / 4 x π x 
I = 3.501 x W/
Therefore the intensity of the light from the bulb would be
3.501 x W/
Answer: False but read why
Heated water molecules expands ( become less dense) but they do float to the top.
Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
Answer:
a) A = 4.0 m
, b) w = 3.0 rad / s
, c) f = 0.477 Hz
, d) T = 20.94 s
Explanation:
The equation that describes the oscillatory motion is
x = A cos (wt + fi)
In the exercise we are told that the expression is
x = 4.0 cos (3.0 t + 0.10)
let's answer the different questions
a) the amplitude is
A = 4.0 m
b) the frequency or angular velocity
w = 3.0 rad / s
c) angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 3 / 2π
f = 0.477 Hz
d) the period
frequency and period are related
T = 1 / f
T = 1 / 0.477
T = 20.94 s
e) the phase constant
Ф = 0.10 rad
f) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф)
speed is maximum when sine is + -1
v = A w
v = 4 3
v = 12 m / s
g) the angular velocity is
w² = k / m
k = m w²
k = 1.2 3²
k = 10.8 N / m
h) the total energy of the oscillator is
Em = ½ k A²
Em = ½ 10.8 4²
Em = 43.2 J
i) the potential energy is
Ke = ½ k x²
for t = 0 x = 4 cos (0 + 0.1)
x = 3.98 m
j) kinetic energy
K = ½ m v²
for t = 00.1
²
v = A w sin 0.10
v = 4 3 sin 0.10
v = 1.98 m / s
