Answer:
Answered
Explanation:
x= 0.02 m
E_p= 10.0 J
E_p= 0.5kx^2
10= 0.5k(0.02)^2
solving we get
K= 50.0 N/m
Now
E'_p= 0.5kx'^2
E'_p= 0.5×50×(0.04)^2
E'_p=40 J
b) potential energy is a scalar quantity and it only depends magnitude and not direction so it will remain same in compression and expansion both
c) 20 J = 0.5×50,000×x^2
solving
x= 0.028 m
d) k is 50.0 N/m from above calculation
________b_____ 7 km east
|
| 2km north.
|a
|
°
pythagorean theorem : ✓a² + b² = c²
c² = a² + b² = 4 + 49 = 53
c = ✓53 km
displacement = c = ✓53 km
distance = 10 + 3 + 2 = 15 km
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
Answer:
Work Done = 67.5 J
Explanation:
First we find the value of spring constant (k) using Hooke's Law. Hooke's is formulated as:
F = kx
where,
F = Force Applied = 450 N
k = Spring Constant = ?
x = Stretched Length = 30 cm = 0.3 m
Therefore,
450 N = k(0.3 m)
k = 450 N/0.3 m
k = 1500 N/m
Now, the formula for the work done in stretching the spring is given as:
W = (1/2)kx²
Where,
W = Work done = ?
k = 1500 N/m
x = 70 cm - 40 cm = 0.3 m
Therefore,
W = (1/2)(1500 N/m)(0.3 m)²
<u>W = 67.5 J</u>
