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Anarel [89]
3 years ago
9

For a cosine function with amplitude A=0.75 and period T=10, what is y(4)?

Physics
2 answers:
mixer [17]3 years ago
8 0
 <span>2π/T = 2π/10 = π/5 

y(x) = A sin (wx) = 0.75 sin (πx/5) 

y(4) = 0.75 sin (4π/5) = 0.4408389392... ≈ 0.441</span><span>
</span>
Ainat [17]3 years ago
8 0

Answer:

y(4) = -0.606    

Explanation:

Given that,

The amplitude of cosine function, A = 0.75

Time period, T = 10

The cosine function is given by :

y=A\ cos(\omega t)

Since, \omega=\dfrac{2\pi}{T}

y=A\ cos(\dfrac{2\pi}{T}t)  

y=0.75\ cos(\dfrac{2\pi}{10}t)      

We need to find y (4). Put t = 4 in above equation as :

y(4)=0.75\ cos(\dfrac{2\pi}{10}\times 4)  

y(4) = -0.606       

Hence, this is the required solution.

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2 years ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3
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This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC       Q₃ = 51 nC        Q₂ = 30.5 nC

x₁ = - 1.715m         x₃ = - 1.085m     x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

6 0
3 years ago
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