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3 years ago

Why do we learn useless nonsense school, why don't we learn what we will actually use in life

1 answer:

4 0

I totally agree but, in my opinion its because of the government and what the state has control over. Teacher have little control over it.

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what is the dot product and cross product of of two vectors if the angle is between them is 90 degree?

DaniilM [7]

$\sf{\pink{\underline{\underline{\blue{GIVEN:-}}}}}$

The angle between the two vectors is 90° .

$\sf{\pink{\underline{\underline{\blue{TO\: FIND:-}}}}}$

The dot product of two vectors .
The cross product of two vectors .

$\sf{\pink{\underline{\underline{\blue{SOLUTION:-}}}}}$

⚡ Let $\rm{\vec{a}}$ and $\rm{\vec{b}}$ are the two vectors .

✍️ We have know that,

$\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:.\:\vec{b}\:=\:ab\cos{\theta}\:}}}}$

Where,

θ = 90°

$\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\cos{90^{\degree}}\:}$

cos 90° = 0

$\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\times{0}\:}$

$\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:0\:}$

$\rm{\red{\therefore}}$ [1] The dot product of two vectors is “ 0 ” .

✍️ We have know that,

$\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:\times\:\vec{b}\:=\:ab\sin{\theta}\:}}}}$

Where,

θ = 90°

$\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\sin{90^{\degree}}\:}$

sin 90° = 1

$\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\times{1}\:}$

$\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\:}$

$\rm{\red{\therefore}}$ [2] The cross product of two vectors is “ ab ” .

3 0

2 years ago

Read 2 more answers

What describes why voters xhoose political canidates?

Alexus [3.1K]

The voters political opinions and what they think what is right and wrong.

4 0

3 years ago

Mineral salt compounds (calcium chloride, table salt)

soldier1979 [14.2K]

Answer:

Chloride and Salt

Explanation:

that make up table salt, a.k.a. sodium chloride (NaCl). ... As sodium chloride (NaCl) or calcium chloride (CaCl2) dissolve in water, ... the compound formed when a positive ion combines with a negative ion out of solution, but ... waters may have more if there is weathering or leaching from nearby mineral-rich soils and rocks.

3 0

3 years ago

Interpretar el servicio de la verdad

Lelechka [254]

Umm what are you trying to say

6 0

2 years ago

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut

blsea [12.9K]

Answer:

1. $E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. $E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

$\rho (r) = \alpha (1-\frac{r}{R})$

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

$\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

$\int\, da = 2\pi r h$

where ‘h’ is the length of the imaginary Gaussian surface.

$Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

$Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3. At the boundary where r = R:

$E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}$

As can be seen from above, two E-field values are equal as predicted.

4 0

3 years ago