Answer:
E = 0.01 J
Explanation:
Given that,
The mass of the cart, m = 0.15 kg
The force constant of the spring, k = 3.58 N/m
The amplitude of the oscillations, A = 7.5 cm = 0.075 m
We need to find the total mechanical energy of the system. It can be given by the formula as follows :

Put all the values,

So, the value of total mechanical energy is equal to 0.01 J.
Answer:
Yes, if the system has friction, the final result is affected by the loss of energy.
Explanation:
The result that you are showing is the conservation of mechanical energy between two points in the upper one, the energy is only potential and the lower one is only kinetic.
In the case of some type of friction, the change in energy between the same points is equal to the work of the friction forces
= ΔEm
=
-Em₀
As we can see now there is another quantity and for which the final energy is lower and therefore the final speed would be less than what you found in the case without friction.
=
+ Em₀
Remember that the work of the rubbing force is negative, let's write the work of the rubbing force explicitly, to make it clearer
½ m v² = -fr d + mgh
v = √(-fr d 2/m + 2 gh)
v = √ (2gh - 2fr d/m)
Now it is clear that there is a decrease in the final body speed.
Consequently, if the system has friction, the final result is affected by the loss of energy.
Explanation:
the acceleration will be unchanged according to newton second law of motion
2 near the middle
between ultraviolet and infared
To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

Here,
P = Pressure
d = Diameter
t = Thickness
At the same time the longitudinal stress is given as,

The letters have the same meaning as before.
Then he hoop stress would be,



And the longitudinal stress would be



The Mohr's circle is attached in a image to find the maximum shear stress, which is given as



Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi