Answer:
The stress is 
Explanation:
From the question we are told that
The diameter of the post is 
The length is 
The weight of the loading mass
Generally the radius of the post is mathematically represented as
=> 
Generally the area of the post is
=> 
=> 
Generally the weight exerted by the load is mathematically represented as
=> 
=> 
Generally the stress is mathematically represented as
=> 
=>
Answer:
Bowling Ball: weight on Earth = 49 N
Textbook: Mass = 2 kg; weight on the moon = 3.2 N
Large dog: weight on Earth = 490 N; weight on the moon = 80 N
Law of Universal Gravitation: 
= gravitational force (Newtons/N)
<em>G</em> = gravitational constant, 6.67430 × 10¹¹ 
<em>m</em>₁ and <em>m</em>₂ = masses of two objects (kilograms/kg)
<em>r</em>² = square of distance between centers of the two objects (meters/m)
Have a fantastic day!
Gases do not conduct heat well.
Angstrom = 10^-10 m
for nucleus size are used fermi (femtometer 10^-15 m )
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
