Complete Question:
Given at a point. What is the force per unit area at this point acting normal to the surface with ? Are there any shear stresses acting on this surface?
Answer:
Force per unit area,
There are shear stresses acting on the surface since
Explanation:
equation of the normal,
Traction vector on n,
To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.
If the shear stress, , is calculated and it is not equal to zero, this means there are shear stresses.
Since , there are shear stresses acting on the surface.
Answer:
The original volume of the first bar is half of the original volume of the second bar.
Explanation:
The coefficient of cubic expansivity of substances is given by;
γ = ΔV ÷ (Δθ)
Given: two metal bars with equal change in volume, equal change in temperature.
Let the volume of the first metal bar be represented by , and that of the second by .
Since they have equal change in volume,
Δ = Δ = ΔV
For the first metal bar,
2γ = ΔV ÷ (Δθ)
⇒ Δθ = ΔV ÷ (2γ)
For the second metal bar,
γ = ΔV ÷ (Δθ)
⇒ Δθ = ΔV ÷ (γ)
Since they have equal change in temperature,
Δθ of first bar = Δθ of the second bar
ΔV ÷ (2γ) = ΔV ÷ (γ)
So that;
(1 ÷ 2) = (1 ÷ )
2 =
=
Thus, original volume of the first bar is half of the original volume of the second bar.
Answer:The greater the mass, there greater then gravitational pull on their objects.
There greater the distance,the lower the gravitational pull between the object
Explanation:
Gravitational pull(force) is directly proportional to the products Of The masses. therefore if the mass increase,the gravitational pull(force) also increases.
Gravitational pull(force) is inversely proportional to distance.if the distance between the objects increases,the gravitational pull(force) decreases .