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dlinn [17]
4 years ago
14

Suppose I have a vector that is 7 units long and that makes an angle of +30 degrees from the positive x-axis. I want to add to t

his a vector that is also 7 units long and that makes an angle of 120 degrees from the positive x-axis. What is the correct (x,y) representation of the sum of these two vectors?
Physics
1 answer:
Vinil7 [7]4 years ago
5 0

Answer:

sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

Explanation:

We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis

So x component of the vector =7cos30^{\circ}=7\times 0.866=6.06

y component of the vector =7sin30^{\circ}=7\times 0.5=3.5

So vector will be 6.06i+3.5j

Now other vector of length of 7 units and makes an angle of 120° with positive x-axis

So x component of vector  =7cos120^{\circ}=7\times -0.5=-3.5i

y component of the vector =7sin120^{\circ}=7\times 0.866=6.06j

Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

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4 0
3 years ago
A certain superconducting magnet in the form of a solenoid of length 0.56 m can generate a magnetic field of 6.5 T in its core w
Gnom [1K]

Answer:

The value  is N =36203 \  turns

Explanation:

From the question we are told that

   The length of the solenoid is  l = 0.56 \  m

   The magnetic field is B  =  6.5 \ T

    The current is I = 80 \ A

     The desired temperature is  T = 4.2 \ K

Generally the magnetic field is mathematically represented as

       B  = \frac{\mu_o * N * I }{L }

=>     N = \frac{B  * L }{\mu_o * I }

Here  \mu_o is the permeability of free space with value  

      \mu_o =  4\pi * 10^{-7} N/A^2

So

     N = \frac{6.5  * 0.56 }{ 4\pi * 10^{-7} *  80  }

=>   N =36203 \  turns

6 0
3 years ago
What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240
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Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

As

V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC

For (b) charge density

As charge density σ is given as:

σ=q/(4πR²)

σ=(5.333×10⁻⁹) / (4π×(0.20)²)

σ=10.62 nC/m²

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