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elixir [45]
3 years ago
12

A child’s toy launches a model parachutist of mass 0.40 kg vertically upwards. The model parachutist reaches a maximum height of

8.5 m. Calculate a) the gravitational potential energy gained by the model parachutes, b) the minimum possible speed with which the model parachutist was launched. c) In practice, the launch speed must be greater than the value calculated in (b). Explain why? *
Physics
1 answer:
melisa1 [442]3 years ago
4 0

Answer:

(a) =34.0J

(b) = 13.038m/s

(c) =

Explanation:

(a) mass (m) =0.45kg

Height(h) =8.5m

Gravity (g) =10m/s^2

But P. E (Potential Energy) = mass × gravity × height

P. E = 0.45×10×8.5

P. E = 34.0J

(b) using v^2=u^2 - 2gs

Where v = final velocity

u= initial velocity

g = gravity

s =distance

But at maximum height <u>v</u><u> </u>=0

0^2= u^2 - 2gs

Transpose u^2 we have

u^2 = 2gs

u^2 = 2×10×8.5

u^2 = 170

u = square root of 170

u = 13.038m/s

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Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

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The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

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Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

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outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

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r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

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