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erik [133]
3 years ago
6

In 5 meters, a person running at 0.8 m/s accelerates at 1.6 m/s2. How fast 16 points

Physics
1 answer:
frutty [35]3 years ago
5 0

They were going at a velocity 4.07m/s

<u>Explanation:</u>

Distance s =5 m

initial velocity u= 0.8 m/s

Acceleration a =1.6m/s2

We have to calculate the velocity with which they were going afterwards i.e final velocity.

Use the equation of motion

v^2=u^2+2as\\=0.8^2+2\times 1.6\times 5\\=16.64\\v=4.07 m/s

They were going with a velocity 4.07 m/s afterwards.

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1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x
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Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

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(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

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dQ=\lambda dl (1)⇒Q=\lambda_{0} 2\pi R(2)⇒\lambda_{0}=\frac{Q}{2\pi R}(3)

Applying the technique of charge integration for finite charges:

V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r'  }} \, dQ(4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

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Using the expressions (1),(4) and (5) you obtain:

V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}}  }} \, d\phi

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(B) For the expression of the magnitude of the field E(z), is important to remember:

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But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

|E| =-\frac{dV(z)}{dz} (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    } (S_b)

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