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3 years ago

Hey solution that contains all of the salute it can normally hold at a given temperature is

Physics

1 answer:

ankoles [38]3 years ago

8 0

Answer:

A saturated solution

Explanation:

A saturated solution is one that contains the most amount of solute that can be dissolved in it at a given temperature

An example of a saturated solution is carbonated water, which readily gives off bubbles of carbon dioxide gas from areas within the solution to the region above the top surface of the gas in liquid solution

A saturation solution of salt in water can be created by continuing to dissolve salt in a given amount of water until it can no longer dissolve any more salt. However, heating the saturated salt solution, increases the amount of salt that can be dissolved.

Therefore, a solution that contains all of the solute it can normally hold at a given temperature is a saturated solution

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where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed

lesya [120]

Answer:

$E_r(6)=4.35614\ MPa$

Explanation:

$\epsilon$ = Strain = 0.49

$\sigma _0$ = 3.1 MPa

At t = Time = 32 s $\sigma$ = 0.41 MPa

$\tau$ = Time-independent constant

Stress relation with time

$\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)$

at t = 32 s

$0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s$

The time independent constant is 16.0787 s

$E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}$

At t = 6

$\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}$

From the first equation

$\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451$

$E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa$

$E_r(6)=4.35614\ MPa$

6 0

3 years ago

a professional baseball player can pitch a baseball with a velocity of 44.7m/s towards home plate. If a baseball weighs 1.4 N, h

kumpel [21]

1.4 N is a weight so calculating it's mass
1.4/9.8 = 0.1428 kg
momentum will be 0.1428*44.7 = 6.38 kgm/s

3 0

3 years ago

Read 2 more answers

A man rides up in an elevator at 12 m. He gains 6500 J of gravitational potential energy. what is the man's mass?

iren2701 [21]

Answer:

We know that potential energy of a body;

= mass(m)× gravitational acceleration(g) × height(h)

Lets find out the mass of the body

P.E. = mgh

=> 6500J = mass × 9.8m/s^2 × 12m

=>6500J = mass × ( 9.8 × 12 ) × ( m/s^2 × m)

=> 6500 Nm = m × 117.6 × m^2 / s^2

=> 6500/117.6 Ns^2/m = mass [°.° Ns^2/m = kg]

=> 55.272 Kg = mass

Therefore the mass of the body = 55.272 kg ~ 60 kg (Ans)

Hope it helps you

6 0

3 years ago

19. A person pushes with 6.0 N for 4.0 seconds on a 2.0 kg object.

hram777 [196]

Answer:24NS

Explanation:

Impulse=force x time

Impulse=6 x 4

Impulse=24NS

5 0

4 years ago

On a very muddy football field, a 110kg linebacker tackles an 85kg halfback. Immediately before the collision, the linebacker is

Alchen [17]

Using conservation of energy and momentum you can solve this question. M_l = mass of linebacker
M_ h = mass of halfback
V_l = velocity of linebacker
V_h = velocity of halfback

So for conservation of momentum,
rho = mv

M_l x V_li + M_h x V_hi = M_l x V_lf + M_h x V_hf

For conservation of energy (kinetic)
E_k = 1/2mv^2/ 1/2mV_li^2 + 1/2mV_{hi}^2 = 1/2mV_{lf}^2 + 1/2mV_{hf}^2

Where i and h stand for initial and final values.
We are already told the masses, \[M_l = 110kg\] \[M_h = 85kg\] and the final velocities \[V_{fi} = 8.5ms^{-1}\] and \[V_{ih} = 7.2ms^{-1}

6 0

3 years ago