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BlackZzzverrR [31]
3 years ago
6

vector A makes equal angles with x,y and z axis. value of its components (in terms of magnitude of vector A will be?

Physics
2 answers:
kiruha [24]3 years ago
5 0
X^2+y^2+z^2=A^2
But here XY and Z are all equal so
3X^2=A^2
X=A/(sqrt(3))
Each component is the value of a divided by the square root of three. This way if you square then and add them up it equals a squared
djverab [1.8K]3 years ago
5 0

Answer:

All three components are

A_x = \frac{A}{\sqrt3}

A_y = \frac{A}{\sqrt3}

A_z = \frac{A}{\sqrt3}

Explanation:

As we know that sum of all three components of the vector will give us resultant vector

So here we can say that

A^2 = A_x^2 + A_y^2 + A_z^2

since it is given that all three components are of same magnitudes so

A_x = A_y = A_z

now we have

A^2 = 3A_x^2

so we have

A_x = A_y = A_z = \frac{A}{\sqrt3}

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A laser of wavelength 720 nm illuminates a double slit where the separation between the slits is 0.22 mm. Fringes are seen on a
kumpel [21]

Answer:

The appropriate solution is "2.78 mm".

Explanation:

Given:

\lambda = 720 \ nm

or,

  = 720\times 10^{-9} \ m

D=0.85 \ m

d = 0.22 \ mm

or,

  =0.22 \times 10^{-3} \ m

As we know,

Fringe width is:

⇒ \beta=\frac{\lambda D}{d}

hence,

Separation between second and third bright fringes will be:

⇒ \theta=\beta=\frac{\lambda D}{d}

       =\frac{720\times 10^{-9}\times 0.85}{0.22\times 10^{-3}}

       =2.78\times 10^{-3} \ m

or,

       =2.78 \ mm

8 0
2 years ago
A ball rolls down an incline with an acceleration of 10 cm/s^2. If it starts with an initial velocity of 0 cm/s and has a velocy
11Alexandr11 [23.1K]
Given a = 10 cm/s²
          u = 0 cm/s
          v = 50 cm/s
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8 0
3 years ago
The magnetic field at the center of a 1.50-cm-diameter loop is 2.70 mT . Part A. What is the current in the loop?
elixir [45]

Explanation:

It is given that,

Diameter of the circular loop, d = 1.5 cm

Radius of the circular loop, r = 0.0075 m

Magnetic field, B=2.7\ mT=2.7\times 10^{-3}\ T

(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :

B=\dfrac{\mu_o I}{2r}

I=\dfrac{2Br}{\mu_o}

I=\dfrac{2\times 2.7\times 10^{-3}\times 0.0075}{4\pi \times 10^{-7}}

I = 32.22 A

(b) The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

r=\dfrac{\mu_o I}{2\pi B}

r=\dfrac{4\pi \times 10^{-7}\times 32.22}{2\pi \times 2.7\times 10^{-3}}

r = 0.00238 m

r=2.38\times 10^{-3}\ m

Hence, this is the required solution.

8 0
3 years ago
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