Answer:
u" + 40u' + 49u = 2 sin(t/6)
upp + 40up + 49u = 2 sin(t/6)
Explanation:
Step 1: Data given
mass = 5 kg
L = 20 cm = 0.2 m
F = 10 sin(t/6)N
Fd(t) = - 6 N
u(0) = 0.03 m/s
u(0) = 0
u'(0) = 3 cm/s
Step 2:
ω =kL
k = ω/L = m*g /L = (5*9.8)/0.2 = 245 kg/s²
Since Fd(t) = -γu'(t) we know:
γ =- Fd(t) / u'(t) = 6N/ 0.03 m/s = 200 Ns/m
The initial value problem which describes the motion of the mass is given by
5u" + 200u' + 245u = 10 sin(t/6) u(0) = 0 ; u'(0) = 0.03
This is equivalent to:
u" + 40u' + 49u = 2 sin(t/6) u(0) = 0 ; u'(0) = 0.03
upp + 40up + 49u = 2 sin(t/6)
With u in m and t in s
Answer:
B
Explanation:
Transformation of energy involves conversion of energy from one form to another for example our movement around involves the conversion of chemical energy stored in the food we eat to other forms of energy such as kinetic energy for the movement, electrical energy in the neurons for impulses and others
The ball posses gravitational potential energy since it is held at a displacement to the ground ( zero point) and when released, the gravitational potential energy is converted to kinetic energy which leads to the fall of the ball until it is at zero displacement to the earth. The board likewise when bent to its maximum extent stored elastic potential energy as a result of the partial displacement of its constituent particle provided it is not stretch beyond its elastic limit which can lead to deformation of the board and the elastic potential energy lost.
An electron has a negative charge. Hope this helps.
<span>The correct frequency when you tune a guitar is
when you hear the right tune in your own hearing and standard. The measure
frequency of a guitar string is when you measure the tune of the string
correctly. This is not the same because manual tuning is affected by many
factors.</span>
Answer:
The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R
Explanation:
Given the data in the question;
Using the Clapeyron equation
where 
is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu
T is the temperature ( 15 + 460 )R
m is the mass of water ( 0.5 Ibm )
is specific volume ( 1.5 ft³ )
we substitute
/ 
272.98 Ibf-ft²/R
Now,
where P₁ is the initial pressure ( 50 psia )
P₂ is the final pressure ( 60 psia )
T₁ is the initial temperature ( 15 + 460 )R
T₂ is the final temperature = ?
we substitute;
480.275 R
Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R
