The answer is the third statement- Tundra, wetland, rainforest, desert
Answer:
γ
=0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ = displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ
By comapring both equations, we get
P/A = Gγ ------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN
Answer:
Quando um (ou todos) os outros coeficientes de uma equação do segundo grau são iguais a zero, essa equação é chamada incompleta. Neste artigo, analisaremos os métodos que podem ser usados para resolver equações incompletas, no caso em que o coeficiente C = 0, ou seja, o coeficiente é nulo.
Explanation:
Um lado da equação será a potência e outro, o número inteiro. De outro modo, transforme a equação deixando-a isolada em um dos lados. Reescreva a equação. Prepare-a a fim de extrair o logaritmo de ambos os lados, que é o inverso da potência. Você pode calcular o logaritmo de base.
Answer:
0.506N
Explanation:
In this question, we are asked to calculate the total drag force on a plate which is oriented parallel to an air flow at a particular temperature and atmospheric pressure.
Please check attachment for complete solution, plate diagram and step-by-step explanation
Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.
