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Ludmilka [50]
3 years ago
8

Two paramedics are lifting a person on a stretcher. One of the paramedics exerts a force of 350 N at 58o above the horizontal an

d the other exerts a force of 410 N at 43o above the horizontal. What is the weight of the person and the stretcher?
Physics
2 answers:
lesantik [10]3 years ago
4 0

 F1 = 350 N
 q1 = 58.0°
 F2 = 410 N
q2 = 43°
 Fy,net = F1(sin q1) + F2(sin q2) = (350 N)(sin 58°) + (410 N)(sin 43)
 Fy,net = 3.0 × 102 N + 2.8 × 102 N Fy,net = 580 N
cupoosta [38]3 years ago
3 0
  <span> Fy = Fy1 + Fy2
Fy = 350*sin(58) + 410*sin(43)
Fy = 576.44 N </span>
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Nana76 [90]

Answer:

Electron shell

Nucleus

Neutrons

Explanation:

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3 years ago
Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an
castortr0y [4]

Answer:

  • Fx = -9.15 N
  • Fy = 1.72 N
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Explanation:

You apparently want the sum of forces ...

  F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

  f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

  = -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

  ≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

  ≈ (-9.15309, 1.71982)

The resultant component forces are ...

  • Fx = -9.15 N
  • Fy = 1.72 N

Then the magnitude and direction of the resultant are

  F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

  F∠γ ≈ 9.31∠-10.6°

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