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Natalija [7]
3 years ago
15

If a sheet of aluminum foil (d = 2.70 g/cm3) weighs 5.31 lbs and is 0.0130 mm thick, the surface area of the foil is closest toâ

¦?
Physics
1 answer:
Salsk061 [2.6K]3 years ago
5 0
Given:
d = 2.70 g/cm³, the density
M = 5.31 lbs, the mass
t = 0.0130 mm, thickness

Convert all data into SI units.
d = (2.70 g/cm³)*(1/1000 kg/g)*(10² cm/m)³
   = 2.7 x 10³ kg/m³
M = (5.31 lb)*(0.543592 kg/lb)
    = 2.4086 kg
t = (0.013 mm)*(10⁻³ m/mm)
  = 1.3 x 10⁻⁴ m

Let the surface area f the foil be A m².
Then the mass of the foil is 
M = d*A*t
    = (2.7 x 10³ kg/m³)*(A m²)*(1.3 x 10⁻⁴ m) 
    = 0.351A kg

Because the mass is 2.4086 kg, therefore
A = 2.4086/0.351
   = 6.862 m²

Answer: The surface area is 6.862 m².

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three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25
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The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

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2 years ago
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