Answer:
5J/mol
Explanation:
H-H eclipsed = Torsional strain = 4.0 kJ/mol
H-CH3 eclipsed = Mostly torsional strain = 6.0 kJ/mol
CH3-CH3 eclipsed =Torsional and steric strain = 11.0 kJ/mol
CH3-CH3 gauche = Steric strain = 3.8 kJ/mol
You know each H-H eclipsed is 4 kJ/mol (8 kJ/mol total) and the reported barrier is 13 kJ/mol. Therefore the H-I eclipsed interaction must be 5 kJ/mol, (13-8 = 5).
<span>Assume: neglect of the collar dimensions.
Ď_h=(P*r)/t=(5*125)/8=78.125 MPa ,Ď_a=Ď_h/2=39 MPa
τ=(S*Q)/(I*b)=(40*〖10〗^3*π(〖0.125〗^2-〖0.117〗^2 )*121*〖10〗^(-3))/(π/2 (〖0.125〗^4-〖0.117〗^4 )*8*〖10〗^(-3) )=41.277 MPa
@ Point K:
Ď_z=(+M*c)/I=(40*0.6*121*〖10〗^(-3))/(8.914*〖10〗^(-5) )=32.6 MPa
Using Mohr Circle:
Ď_max=(Ď_h+Ď_a)/2+âš(Ď„^2+((Ď_h-Ď_a)/2)^2 )
Ď_max=104.2 MPa, Ď„_max=45.62 MPa</span>
v1f = -0.16 ms
Explanation:
Use the conservation law of linear momentum:
m1v1i + m2v2i = m1v1f + m2v2f
where
v1i = v2i = 0
m1 = 160 kg
m2 = 0.50 kg
v2f = 50m/s
v1f = ?
So we have
0 = (160 kg)v1f + (0.5 kg)(50 m/s)
v1f = -(25 kg-m/s)/(160 kg)
= -0.16 m/s
Note: the negative sign means that its direction is opposite that of the arrow.
<span>A virus particle is not a complete cell but an intracellular parasite. Hence, it cannot reproduce without the help of a host cell. Once inside a host cell, the virus is made in such a way that it replicates itself. There are two ways in which viruses reproduce or multiply their numbers: the lytic cycle and the lysogenic cycle. </span>
Answer:
100 Joules
Explanation:
Applying,
W = mgh................... Equation 1
Where W = workdone to hold the box above the ground, mg = weight of the box, h = height of the box.
From the question,
Given: mg = 10 newtons, h = 10 meters.
Substitute these values into equation 1
W = 10×10
W = 100 Joules.
Hence the amount of workdone is 100 Joules
