Which SPF level should a person look for in sunscreen as part of planning a day at the beach? pls i need it now

A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformati

Slav-nsk [51]

The question is incomplete. The complete question is :

A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformation cycle. At a time, t = 0, a tensile stress of 20 MPa is applied instantaneously and maintained for 100 s. The stress is then removed at a rate of 0.2 MPa s−1 until the polymer is unloaded. If the creep compliance of the material is given by:

J(t) = Jo (1 - exp (-t/to))

Where,

Jo= 3m^2/ GPA

to= 200s

Determine

a) the strain after 100's (before stress is reversed)

b) the residual strain when stress falls to zero.

Answer:

a)-60GPA

b) 0

Explanation:

Given t= 0,

σ = 20Mpa

Change in σ= 0.2Mpas^-1

For creep compliance material,

J(t) = Jo (1 - exp (-t/to))

J(t) = 3 (1 - exp (-0/100))= 3m^2/Gpa

a) t= 100s

E(t)= ΔσJ (t - Jo)

= 0.2 × 3 ( 100 - 200 )

= 0.6 (-100)

= - 60 GPA

Residual strain, σ= 0

E(t)= Jσ (Jo) ∫t (t - Jo) dt

3 × 0 × 200 ∫t (t - Jo) dt

E(t) = 0

5 0

3 years ago

Milk containing 3.7% fat and 12.8% total solids is to be evaporated to produce a product containing 7.9% fat. What is the yield

ratelena [41]

Answer:

the yield of product is YP=46.835 % and the concentration of solids is

Cs = 27.33%

Explanation:

Assuming that all the solids and fats remains in the milk after the evaporation, then the mass of product mP will be

Mass of fat in 100 kg of milk = 100 kg* 0.037 = mP* 0.079

mP = 100 kg* 0.037/0.079 = 46.835 kg

then the yield YP of the product is

YP= mP / 100 kg = 46.835 kg / 100 kg = 46.835 %

YP= 46.835 %

the concentration of solids Cs is

Mass of solids in 100 kg of milk = 100 kg* 0.128 = 46.835 kg * Cs

Cs = 100 kg* 0.128 / 46.835 kg = 0.2733 = 27.33%

Cs = 27.33%

3 0

3 years ago

A speed boat increases its speed uniformly from vi = 20.0 m/s to vf = 30.0 m/s in a distance of 2.00 x 10^2m. (a) Draw a coordin

pychu [463]

a) See graph in attachment

b) The suvat equation to use is $v_f^2 - v_i^2 = 2as$

c) The acceleration is $a=\frac{v_f^2-v_i^2}{2s}$

d) The acceleration is $1.25 m/s^2$

e) The time needed is 8 s

Explanation:

a)

For this part, find in attachment the diagram representing this situation.

Since we are not given any particular direction for the motion, we choose the x-direction as the direction of motion of the boat.

Then we have the following:

- The initial position of the boat is $x_i = 0$ , the origin

- The final position of the boat is $x_f = 200 m$

- The initial velocity of the boat is $v_i = 20.0 m/s$

- The final velocity of the boat is $v_f = 30.0 m/s$

Note that the arrow representing the final velocity is longer than that of the initial velocity, since the final velocity is larger.

b)

The motion of the speed boat is a uniformly accelerated motion (motion at constant acceleration), therefore we can use one of the suvat equations. In this particular problem, we know the following quantities:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat