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4 years ago

When the reaction:

1 answer:

3 0

The last sentence is a little vague.
However, I will assume that by the last sentence you meant
$Reaction~Rate=[CH_3Cl][H_2O]$
The reaction order respective to any substance is the value of the exponent on that substance in the Reaction Rate equation.
We see that the exponent on both the substance is 1 so the reaction order with respect to each substance is 1.

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A thin plate 6 ft long and 3 ft wide is submerged and held stationary in a stream of water (T = 60°F) that has a velocity of 17

VashaNatasha [74]

A) In the case of the Boundary Thickness Layer we use the given formula,

$\delta = \frac{4.91x}{\sqrt{Re}}$

We know as well that,

Re = Número de Reynolds = $\frac{U*x}{\upsilon}$

Where,

U = velocity

$\upsilon$ = kinematic viscosity

For water, kinematic viscosity, $\upsilon = 1.21*10^{-5} ft^2 /s$

So, $500,000 = \frac{ 17x}{(1.21*10^{-5})}$

$x = 0.355 ft$

$d = \frac{4.91*0.355}{\sqrt {500000}}$

$d = 0.002465 ft = 0.029in$

B) For flat plate boundary layer. Given the Critical Reynolds Number.= 5*10^5 we know that is equal to Re above.

Thus, $x = 0.355 ft$

C. Wall shear stress,

$\tau = \mu*\sqrt{ U^3 / (2*\nu*x) }$

For water, dynamic viscosity, $\nu$ = 2.344*10^-5 lbf-s/ft^2

$\tau = 2.344*10^-5 \sqrt {17^3 / (2*1.21*10^{-5}*0.355)}$

$\tau = 0.5605 lbf/ft^2$

4 0

3 years ago

A student walks to the right 25-m along the 800 hall in 15-s. They turn around and walk 15-m to the left in 8.0-s. Calculate the

Ludmilka [50]

Answer:

Explanation:

Total distance covered = 25 + 15 = 40 m

Total time = 15 + 8 = 23 s

Average speed = total distance covered / total time

= 40 / 23

= 1.74 m / s

Total displacement = 25 - 15 = 10 m

Total time = 15 + 8 = 23 s

Average velocity = total displacement / total time

= 10 / 23

= .434 m / s to the right .

3 0

4 years ago

The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5

KiRa [710]

Answer:

0.68 s

Explanation:

We are given that

Initial velocity of box= $u=13m/s$

Final velocity of box=v=11.5 m/s

Distance=d=8.5 m

We have to find the time taken by box to slow by this amount.

We know that

$v^2-u^2=2as$

Substitute the values

$(11.5)^2-(13)^2=2a(8.5)$

$132.25-169=17a$

$-36.75=17a$

$a=\frac{-36.75}{17}=-2.2m/s^2$

We know that

Acceleration= $a=\frac{v-u}{t}$

Substitute the values

$-2.2=\frac{11.5-13}{t}$

$-2.2=\frac{-1.5}{t}$

$t=\frac{1.5}{2.2}=0.68 s$

Hence, the time taken by box to slow by this amount=0.68 s

8 0

4 years ago

Two identical 0.400 kg masses are pressed against opposite ends of a light spring of force constant 1.75 N/cm, compressing the s

nlexa [21]

Answer:

0.853 m/s

Explanation:

Total energy stored in the spring = Total kinetic energy of the masses.

1/2ke² = 1/2m'v².................... Equation 1

Where k = spring constant of the spring, e = extension, m' = total mass, v = speed of the masses.

make v the subject of the equation,

v = e[√(k/m')].................... Equation 2

Given: e = 39 cm = 0.39 m, m' = 0.4+0.4 = 0.8 kg, k = 1.75 N/cm = 175 N/m.

Substitute into equation 2

v = 0.39[√(1.75/0.8)

v = 0.39[2.1875]

v = 0.853 m/s

Hence the speed of each mass = 0.853 m/s

7 0

3 years ago

A train is moving in a straight railway where it covered one third of the distance with

Alexxandr [17]

Answer:

<em>The average speed of the train is 45 km/h</em>

Explanation:

<u>Speed</u>

It's defined as the distance (d) per unit of time (t) traveled by an object. The formula is:

$\displaystyle v=\frac{d}{t}$

Let's call x the total distance covered by the train. It covered d1=1/3x with a speed of v1=25 km/h. The time taken is calculated solving for t:

$\displaystyle t_1=\frac{d_1}{v_1}$

$\displaystyle t_1=\frac{1/3x}{25}$

$\displaystyle t_1=\frac{x}{75}$

Now the rest of the distance:

d2 = x - 1/3x = 2/3x

Was covered at v2=75 km/h. Thus the time taken is:

$\displaystyle t_2=\frac{d_2}{v_2}$

$\displaystyle t_2=\frac{2/3x}{75}$

$\displaystyle t_2=\frac{2x}{225}$

The total time is:

$\displaystyle t_t=\frac{x}{75}+\frac{2x}{225}$

$\displaystyle t_t=\frac{3x}{225}+\frac{2x}{225}$

$\displaystyle t_t=\frac{5x}{225}$

Simplifying:

$\displaystyle t_t=\frac{x}{45}$

The average speed is the total distance divided by the total time:

$\displaystyle \bar v=\frac{x}{\frac{x}{45}}$

Simplifying:

$\boxed{\displaystyle \bar v=45\ km/h}$

The average speed of the train is 45 km/h

5 0

3 years ago