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2 years ago

) The magnitude of scalar product of two unit vectors perpendicular to each other is

Physics

1 answer:

fenix001 [56]2 years ago

5 0

This question involves the concept of the scalar product.

The magnitude of the scalar product will be "0".

<h3>SCALAR PRODUCT</h3>

The scalar product, also known as the dot product of the two vectors is given by the following formula:

$A.B = |A||B|Cos\theta$

where,

A.B = Scalar product = ?
|A| = Mangnitude of vector A = 1 unit
|B| = Magnitude of Vector B = 1 unit
θ = Angle between vectors = 90°

Therefore,

$A.B = (1)(1)Cos90^o = (1)(1)(0)$

A.B = 0

Learn more about scalar product here:

brainly.com/question/6849226

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Explanation:

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3 years ago

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,

viktelen [127]

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
$=3.67\cdot 10^{-11}F$

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$ . We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

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3 years ago

How much pressure is created when you apply a

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Pressure
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2 years ago

How does the direction of the electric current, moving across a battery powered device, differ from the direction of travel in t

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3 years ago

Read 2 more answers

1. Two forces act on a box as follows: F1 = 100 N at 01 = 170° and F2 = 75 N

lilavasa [31]

Answer:

a) F = 64.30 N, b) θ = 121.4º

Explanation:

Forces are vector quantities so one of the best methods to add them is to decompose each force and add the components

let's use trigonometry

Force F1

sin 170 = F_{1y} / F₁

cos 170 = F₁ₓ / F₁

F_{1y} = F₁ sin 170

F₁ₓ = F₁ cos 170

F_{1y} = 100 sin 170 = 17.36 N

F₁ₓ = 100 cos 170 = -98.48 N

Force F2

sin 30 = F_{2y} / F₂

cos 30 = F₂ₓ / F₂

F_{2y} = F₂ sin 30

F₂ₓ = F₂ cos 30

F_{2y} = 75 sin 30 = 37.5 N

F₂ₓ = 75 cos 30 = 64.95 N

the resultant force is

X axis

Fₓ = F₁ₓ + F₂ₓ

Fₓ = -98.48 +64.95

Fₓ = -33.53 N

Y axis

F_y = F_{1y} + F_{2y}

F_y = 17.36 + 37.5

F_y = 54.86 N

a) the magnitude of the resultant vector

let's use Pythagoras' theorem

F = Ra Fx ^ 2 + Fy²

F = Ra 33.53² + 54.86²

F = 64.30 N

b) the direction of the resultant

let's use trigonometry

tan θ’= F_y / Fₓ

θ'= $tan^{-1} \frac{F_y}{F_x}$

θ'= tan⁻¹ (54.86 / (33.53)

θ’= 58.6º

this angle is in the second quadrant

The angle measured from the positive side of the x-axis is

θ = 180 -θ'

θ = 180- 58.6

θ = 121.4º

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2 years ago

) The magnitude of scalar product of two unit vectors perpendicular to each other is ​

) The magnitude of scalar product of two unit vectors perpendicular to each other is