All categories

3 years ago

Biomedical, electrical and civil engineering essay

Engineering

1 answer:

ArbitrLikvidat [17]3 years ago

3 0

Answer:

7th A good time for you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my replying sooner than later this month to you arshpreet May God bless you live a long life of shapes in which class do read it and I am asking you are nagre andg my

You might be interested in

Which of the following maintenance items helps to ensure the vehicles engine lasts as long as possible?

Paladinen [302]

Answer:

because oil is preventing corrosion and rust

8 0

2 years ago

In C++ the declaration of floating point variables starts with the type name float or double, followed by the name of the variab

Aliun [14]

Answer:

The given grammar is :

S = T V ;

V = C X

X = , V | ε

T = float | double

C = z | w

Nullable variables are the variables which generate ε ( epsilon ) after one or more steps.

From the given grammar,

Nullable variable is X as it generates ε ( epsilon ) in the production rule : X -> ε.

No other variables generate variable X or ε.

So, only variable X is nullable.

First of nullable variable X is First (X ) = , and ε (epsilon).

L.H.S.

The first of other varibles are :

First (S) = {float, double }

First (T) = {float, double }

First (V) = {z, w}

First (C) = {z, w}

R.H.S.

First (T V ; ) = {float, double }

First ( C X ) = {z, w}

First (, V) = ,

First ( ε ) = ε

First (float) = float

First (double) = double

First (z) = z

First (w) = w

Explanation:

7 0

3 years ago

Please help will give brainliest please answer all 3

larisa [96]

Answer: For #1 I'm going to go with A because that has to do with biology

For #2 I'm going to go with B oceans because that has to do with plant life (and life in general).

For #3 I'll say marine/maritime engineer (you can just say marine)

Hope it helps!

8 0

3 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical

stealth61 [152]

Answer:

Explanation:

We know that Drag force $F_D$

$F_D=\dfrac{1}{2}C_D\rho AV^2$

Where

$C_D$ is the drag force constant.

A is the projected area.

V is the velocity.

ρ is the density of fluid.

Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.

Example:

When we place our hand out of the window in a moving car ,we feel a force in the opposite direction and feel like some one trying to pull our hand .This pulling force is nothing but it is drag force.

6 0

3 years ago