Convection is the transfer of heat energy by:

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

4 years ago

When you speak or sing, you force air through your ?

zaharov [31]

You force air from your lungs up through your larynx which is commonly called the voicebox

6 0

3 years ago

A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3s later. How high

Marysya12 [62]

We use kinematic equation,

$h=ut+\frac{1}{2} gt^2$

Here, h is vertical height, u is initial vertical velocity, t is time taken and g is acceleration due to gravity.

As diver dives out horizontally, his velocity is directed horizontally; that is, the initial vertical velocity is 0. So above equation becomes

$h=\frac{1}{2} gt^2$

Given, t =3 s.

Therefore,

$h=\frac{1}{2}\times 9.8m/s^2(3\ s)^2 =0.5\times 9.8 m/s^2\times 9 s^2\\\\h=44.1\ m$ .

Now the horizontal distance of the diver to hit the water from base,

$=1.8\ m/s \times 3\ s=5.4\ m$

6 0

4 years ago

What is the answer to {36m= ? dm}

Andrew [12]

Answer:

360

Explanation:

just add a 0 when using dm

8 0

3 years ago

A uniform cylinder of radius R and mass M is mounted so as to rotate freely about a horizontal axis that is parallel to, and a d

almond37 [142]

Answer:

I = M*(0.5R^2 + h^2)

w = sqrt [2*g*h / (0.5R^2 + h^2)]

Explanation:

Given:

- The radius of the cylinder R

- The mass of the cylinder M

- The distance between horizontal axis h

- The moment of inertia of cylinder I_c

Solution:

- When the axis of rotation and axis with respect to rotation inertia is required do not coincide then we have to apply parallel axis theorem to calculate the moment of inertia about the required axis which is at a distance h from body rotational axis as follows:

I = I_c + M*h^2

- Where I_c of a cylinder is = 0.5*M*R^2

I = 0.5*M*R^2 + M*h^2

I = M*(0.5R^2 + h^2)

- From conservation of total mechanical energy of the cylinder, the change in gravitational potential energy ΔP.E plus the change in kinetic energy ΔK.E of the cylinder must be zero:

ΔP.E = ΔK.E

M*g*h = 0.5*I*w^2

M*g*h = 0.5*M*(0.5R^2 + h^2)*w^2

w^2 = 2*g*h / (0.5R^2 + h^2)

w = sqrt [2*g*h / (0.5R^2 + h^2)]

Where,

w: The angular speed of the cylinder

8 0

3 years ago