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Nataly_w [17]
3 years ago
5

At a distance r1 from a point charge, the magnitude of the electric field created by the charge is 226 N/C. At a distance r2 fro

m the charge, the field has a magnitude of 134 N/C. Find the ratio r2/r1.

Physics
2 answers:
jenyasd209 [6]3 years ago
8 0

Explanation:

Below is an attachment containing the solution.

Svetradugi [14.3K]3 years ago
6 0

Answer:

r2/r1 = 1.3

Explanation:

Electric field is given as:

E = kq/r²

At a distance r1,

226 = kq/(r1)² - - - - - - - - - - - - - (1)

At a distance r2,

134 = kq/(r2)² - - - - - - - - - - - - - (2)

From (1),

kq = 226 * (r1)²

From (2),

kq = 134 * (r2)²

Equating and then solving,

134 * (r2)² = 226 * (r1)²

(r2)²/(r1)² = 226/134

(r2)²/(r1)² = 1.687

=> r2/r1 = 1.3

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Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

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Explanation:

Given the data in the question;

V = 20 m/s

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drag co-efficient CD = 1.855

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a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

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FD = 742 N

we know that;

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Fg = gravity force

Fg = mg

so

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

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Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

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Now, FD = 1/2 × CD × β × A × V², we solve for A

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we substitute

A = 1780.8  / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

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Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

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we know that;

FD = Fg = M"g

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from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

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Answer:

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