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stira [4]
3 years ago
11

A startled armadillo leaps upward, rising 0.532 m in the first 0.202 s. (a) What is its initial speed as it leaves the ground? (

b) What is its speed at the height of 0.532 m? (c) How much higher does it go? Use g=9.81 m/s2.
Physics
1 answer:
MrMuchimi3 years ago
7 0
A) d=v_0t+\frac{1}{2}at^2
0.532 m = 0.202v_0 + ½(-9.81 m/s²)(0.202 s)²
v_0 = 3.62 m/s

B) d = \frac{1}{2}(v+v_0)t
0.532 m = ½(v + 3.62 m/s)(0.202) s
v = 1.65 m/s

C) v^2=v_0^2+2ad
At the highest point v = 0, so
0 = (1.65 m/s)^2 + 2(-9.81 m/s²)d
d = 0.139 m
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Explanation:

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With k the Coulomb constant, q the charge and r the distance of the charge to the test point

       Et = E1 + E2 + E3

       E1 = k q / (x-a)²

       E2 = k (-2q) / x²  

       E3 = k q / (x + a)²

       Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ  where x << 1, for this we take factor like x from all the equations

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We use binomial expansion

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     (1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

       

   Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

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Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

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