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
There are two types of electric charges; positive and negative
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- hope this helps
Answer:
217.43298 m/s
Explanation:
= Mass of bullet = 19 g
= Mass of bob = 1.3 kg
L = Length of pendulum = 2.3 m
= Angle of deflection = 60°
u = Velocity of bullet
Combined velocity of bullet and bob is given by
As the momentum is conserved
The speed of the bullet is 217.43298 m/s
Answer:
Workdone = 1960 Joules.
Explanation:
Given the following data;
Mass = 5kg
Force = 49N
Height (distance) = 40m
To find the workdone;
Workdone = force * distance
Substituting into the equation, we have;
Workdone = 49*40
Workdone = 1960 Joules.
Therefore, the amount of work done on the bowling ball to lift it is 1960 Joules.
Answer:
The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²
Explanation:
Thickness of the wall is L= 20cm = 0.2m
Thermal conductivity of the wall is K = 2.79 W/m·K
Temperature at the left side surface is T₁ = 50°C
Temperature of the air is T = 22°C
Convection heat transfer coefficient is h = 15 W/m2·K
Heat conduction process through wall is equal to the heat convection process so
Expression for the heat conduction process is
Expression for the heat convection process is
Substitute the expressions of conduction and convection in equation above
Substitute the values in above equation
Now heat flux through the wall can be calculated as
Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²
