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sergiy2304 [10]
3 years ago
12

A vector has an x-component of length 10 and a y-component of length 3. What is the angle of the vector? (Hint: Use the inverse

tangent.)
Physics
1 answer:
Sonja [21]3 years ago
5 0
The vector, the x-component and the y-component form a rectangle triangle where the vector is the hypothenuse and the x and y components are the two sides.
Calling \alpha the angle between the vector and the horizontal direction (x), the two sides are related to \alpha by
\tan \alpha  =  \frac{v_y}{v_x}
where vy and vx are the two components on the y- and x-axis. Using vx=10 and vy=3 we find
\tan \alpha  =  \frac{3}{10} =0.3
And so the angle is
\alpha = \arctan (0.3)=16.7^{\circ}
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What did Bohr hypothesize about the atom?
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7 0
4 years ago
A particle of charge +2q is placed at the origin and particle of charge -q is placed on the x-axis at x = 2a. Where on the x-axi
zzz [600]

Answer:

r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}  

Explanation:

We know that the electric force equation is:

F=k\frac{q_{1}*q_{2}}{r^{2}}

  • k is the electric constant 9*10^{9} Nm^{2}/C^{2}
  • r is the distance between the particles
  • q1 and q2 are the particle

Now, we have three particles, the first one at x=0, the second one at x=2a and the third in some place between these two particle.

1. Let's find the electric force between the first particle and the third particle.

F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}

F_{31}=k\frac{q*2q}{r_{31}^{2}}

F_{31}=k\frac{2q^{2}}{r_{31}^{2}}

r(31) is the distance between 3 and 1

2. Now,  let's find the electric force between the third particle and the second particle.

F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}

F_{32}=k\frac{q*(-q)}{r_{32}^{2}}

F_{32}=-k\frac{q^{2}}{r_{32}^{2}}

r(32) is the distance between 3 and 2.

Now, r_{31}+r_{32}=2a or r_{32}=2a-r_{31}

The net force must be zero so:

F_{31}+F_{32}=0[\tex][tex]k\frac{2q^{2}}{r_{31}^{2}}-k\frac{q^{2}}{r_{32}^{2}}=0[\tex]   [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{r_{32}^{2}})=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}})=0[\tex] It means that:[tex]\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}}

We just need to solve it for r(31)

r_{31}^{2}=2(2a-r_{31})^{2}

r_{31}^{2}=2(2a-r_{31})^{2}

r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}  

Therefore the distance from the origin will be:

r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}  

I hope it helps you!        

                 

 

     

4 0
4 years ago
Whats the difference between independent and dependent variables? also, theory and hypothesis?
svlad2 [7]
Independent variable: the variable in an experiment which value is not changed based on any other variable. often denoted by "x"
dependent variable: the variable in an experiment which changes based on the value of the independent variable, ie how much of it you use, how often you use it, etc. often denoted by "y"

theory: an explanation as to why certain phenomena occur. supported and testable, but could change with new research / experiments, etc.
hypothesis: a guess or an estimate of what will happen during an experiment. testable, and may be wrong.

i hope this helps!!
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ad-work [718]

Answer:

heat energy is the form of energy produced by heat

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3 years ago
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