A light bulb, a speaker, and a toaster ovens are all examples of what in an electric circuit.

Find the magnitude of u × v and the unit vector parrallel to u × v in the direction of u × v. u = 2i + 2j - k, v = -i + k

IrinaVladis [17]

Answer:

magnitude = 3

unit vector = $\frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}$

Explanation:

Given vectors:

u = 2i + 2j - k

v = -i + k = -i + 0j + k

(a) u x v is the cross product of u and v, and is given by;

$u X v = \left[\begin{array}{ccc}i&j&k\\2&2&-1\\-1&0&1\end{array}\right]$

u x v = i(2+0) - j(2 - 1) + k(0 - 2)

u x v = 2i - j - 2k

Now the magnitude of u x v is calculated as follows:

| u x v | = $\sqrt{2^2 + (-1)^2 + (-2)^2}$

| u x v | = $\sqrt{4 + 1 + 4}$

| u x v | = $\sqrt{9}$

| u x v | = 3

Therefore, the magnitude of u x v is 3

(b) The unit vector û parallel to u x v in the direction of u x v is given by the ratio of u x v and the magnitude of u x v. i.e

û = $\frac{u X v}{|u X v|}$

u x v = 2i - j - 2k [calculated in (a) above]

|u x v| = 3 [calculated in (a) above]

∴ û = $\frac{2i - j - 2k}{3}$

∴ û = $\frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}$

4 0

3 years ago

How would the amount of carbon dioxide in the atmosphere change if they were no plants?

Flauer [41]

There would be significantly more CO2 in the atmosphere because plants take in CO2 during photosynthesis and fix the carbon into glucose.

8 0

3 years ago

Is the information I’m saying make actual sense to the question?

Arada [10]

Yes you got this buddy you're an A+ student

3 0

4 years ago

Consider a air-filled parallel-plate capacitor with plates of length 8 cm, width 5.52 cm, spaced a distance 1.99 cm apart. Now i

prohojiy [21]

Answer:

The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

Explanation:

Given that,

Length of plates = 8 cm

Width = 5.52 cm

Distance = 1.99 cm

Dielectric constant = 2.6

Length = 4.4 cm

Potential = 0.8 V

We need to calculate the initial capacitance

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times8\times5.52\times10^{-4}}{1.99\times10^{-2}}$

$C=1.96\times10^{-12}$

We need to calculate the final capacitance

Using formula of capacitance

$C'=\dfrac{\epsilon_{0}A_{1}}{d}+\dfrac{k\epsilon_{0}A_{2}}{d}$

Put the value into the formula

$C'=(\dfrac{8.85\times10^{-12}}{1.99\times10^{-2}})((4.4\times5.52)+(3.6\times5.52)2.6)\times10^{-4}$

$C'=3.37\times10^{-12}$

We need to calculate the ratio of the new potential energy to the potential energy before the insertion of the dielectric

Using formula of energy

$\dfrac{E}{E'}=\dfrac{\dfrac{1}{2}CV^2}{\dfrac{1}{2}C'V^2}$

Put the value into the formula

$\dfrac{E}{E'}=\dfrac{1.96\times10^{-12}}{3.37\times10^{-12}}$

$\dfrac{E}{E'}=0.58$

Hence, The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

4 0

4 years ago

....,..................,..,.......

I am Lyosha [343]

I agree, or is this morse code…?

3 0

3 years ago