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vekshin1
3 years ago
7

You guys are amazing :D

Engineering
2 answers:
iren [92.7K]3 years ago
4 0
Thank you. You’re amazing as well !
Sloan [31]3 years ago
3 0
Ik i am thank you tho xoxo
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A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric
diamong [38]

Answer:

Explanation:

Given

q_1=11.5\ nC charge is placed at x=0\ cm

another charge of q_2=-1.2\ nC is at x=3\ cm

We know that Electric field due to positive charge is away  from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to q_2 at some distance r from it

E_2=\frac{kq_2}{r^2}

Now Electric Field due to q_1 is

E_1=\frac{kq_1}{(3+r)^2}

Now E_1+E_2=0

\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0

\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}

\frac{3+r}{r}=3.095

thus r=1.43\ cm

Thus Electric field is zero at some distance r=1.43 cm right of q_2

3 0
3 years ago
5. (5 points) Select ALL statements that are TRUE A. For flows over a flat plate, in the laminar region, the heat transfer coeff
finlep [7]

Answer:

The following statements are true:

A. For flows over a flat plate, in the laminar region, the heat transfer coefficient is decreasing in the flow direction

C. For flows over a flat plate, the transition from laminar to turbulence flow only happens for rough surface

E. In general, turbulence flows have a larger heat transfer coefficient compared to laminar flows 6.

Select ALL statements that are TRUE

B. In the hydrodynamic fully developed region, the mean velocity of the flow becomes constant

D. For internal flows, if Pr>1, the flows become hydrodynamically fully developed before becoming thermally fully developed

Explanation:

7 0
2 years ago
For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value
OleMash [197]

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

Y = 1 - e^{-kt^n}

here Y is percentage of completion  of reaction = 50%

t  is duration of reaction = 146 sec

so,

0.50 = 1 - e^{-k^146^2.1}

0.50 = e^{-k306.6}

taking natural log on both side

ln(0.5) = -k(306.6)

k = 2.26\times 10^{-3}

for 86 % completion

0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}

e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14

-2.26\times 10^{-3} \times t^{2.1} = ln(0.14)

t^{2.1} = 869.96

t = 25.10 sec

5 0
3 years ago
Joe is a chemical engineer whose plant discharges heavy metals into the local river. By the test authorized by the city governme
chubhunter [2.5K]

Answer:

B probably

Explanation:

Because the prompt doesn't specify what sort of violation it could be anything maybe when they release the metals during the day and so on.

5 0
2 years ago
A rocket is launched from rest with a constant upwards acceleration of 18 m/s2. Determine its velocity after 25 seconds
lisabon 2012 [21]

Answer:

The final velocity of the rocket is 450 m/s.

Explanation:

Given;

initial velocity of the rocket, u = 0

constant upward acceleration of the rocket, a = 18 m/s²

time of motion of the rocket, t = 25 s

The final velocity of the rocket is calculated with the following kinematic equation;

v = u + at

where;

v is the final velocity of the rocket after 25 s

Substitute the given values in the equation above;

v = 0 + 18 x 25

v = 450 m/s

Therefore, the final velocity of the rocket is 450 m/s.

5 0
2 years ago
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