Aspecimen of aluminum having a rectangular cross section 9.5 mm × 13.0 mm (0.3740 in. × 0.5118 in.) is pulled in tension with 35600 n (8003 lbf) force, producing only elastic deformation. the elastic modulus for aluminum is 69 gpa (or 10 × 106 psi). calculate the resulting strain.

〖4.1x10〗^(-3)

Explanation:

From the question, the parameters given are:

We can solve this using the formula for elastic modulus ( E ) = stress/strain  

making strain the subject of the formula the equation above becomes strain= stress/ elastic modulus

Where strain= change in length/ length

And  

Stress= force/ area

now let’s calculate value for stress

Stress= 35600/0.0001235=288,259,109  N/m^2

substituting the value for stress into the equation strain= stress/ elastic modulus we have

Strain= 288,259,109 / 69 x 〖10〗^9 = 〖4.1x10〗^(-3)

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)