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Nina [5.8K]
3 years ago
10

Albert dressed as TWO-FACE flips a coin up in the air at an upward velocity of 5.00 m/s. He fails to catch it on the way down an

d it falls down an 8.5 m deep well. Calculate the maximum height and maximum magnitude (could be positive or negative) of velocity of the coin.
Physics
1 answer:
Colt1911 [192]3 years ago
8 0

Answer:

1.28 m, 14 m/s

Explanation:

At the maximum height, the velocity is 0.

Given:

a = -9.8 m/s²

v₀ = 5.00 m/s

v = 0 m/s

x₀ = 0 m

Find:

x

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (5.00 m/s)² + 2(-9.8 m/s²) (x - 0 m)

x = 1.28 m

The maximum speed is at the bottom of the well.

Given:

a = -9.8 m/s²

v₀ = 5.00 m/s

x₀ = 0 m

x = -8.5 m

Find:

v

v² = v₀² + 2a(x - x₀)

v² = (5.00 m/s)² + 2(-9.8 m/s²) (-8.5 m - 0 m)

v = -13.8 m/s

Rounded to 2 sig-figs, the maximum speed is 14 m/s.

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Smaller load:    Work = (5 kg)  x  (9.8 m/s²)  x  (4 m) = 196 joules.

The work required is equal in both cases.

The mass ratio of  2:1  is exactly balanced by
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Answer:

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Explanation:

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Chapter 16, Problem 63. A person is standing in a room at 18 ◦C. The exposed surface area and skin temperature of the person are
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Answer:Q=248.011 W

Explanation:

Given

Temperature of Room T_{\infty }=18^{\circ}\approx 291 K

Area of Person A=1.7 m^2

Temperature of skin T=32^{\circ}\approx 305 K

Heat transfer coefficient h=5 W/m^2.k

Emissivity of the skin and clothes \epsilon =0.9

\Delta T=32-18=14^{\circ}

Total rate of heat transfer=heat Transfer due to Radiation +heat transfer through convection

Heat transfer due radiation Q_1=\epsilon \sigma A(T^4-T_{\infty }^4 )

where \sigma =stefan-boltzman\ constant

Q_1=0.9\times 5.687\times 10^{-8}\times 1.7\times (305^4-291^4)

Q_1=129.01 W

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Q_2=hA(\Delta T)

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. Why are scientific models important? They help scientists prove scientific theories. They help scientists explain complex ques
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The quadriceps muscles pull on the patella simultaneously. Below are the forces from each
Nostrana [21]

Based on the calculation of the resultant of vector forces:

  1. the resultant force due to the quadriceps is 1795 N
  2. the resultant force due to the quadriceps is 1975 N. Training and strengthening the vastus medialis results in a greater force of muscle contraction.

<h3>What is the resultant force due to the quadriceps?</h3>

The resultant of more than two vector forces is given by:

  • F = √Fₓ² + Fₙ²

where:

  • Fₓ is the sum of the horizontal components of the forces
  • Fₙ is the sum of the vertical components of the forces
  • Fx = F₁cosθ + F₂cosθ + F₃cosθ + F₄cosθ
  • Fₙ = F₁sinθ + F₂sinθ + F₃sinθ + F₄sinθ
  • F₁ = 680N, θ = 90 = 30 = 120°
  • F₂ = 220 N, θ = 90 + 16 = 106°
  • F₃ = 600 N, θ = 90 + 15 = 105°
  • F₄ = 480 N, θ = 90 - 35 = 55°

then:

Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 480 * cos 55

Fx = -280.6 N

Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 480 * sin 55

Fₙ = 1773.1 N

then:

F = √(-280.6)² + ( 1773.1)²

F = 1795.16 N

F ≈ 1795 N

Therefore, the resultant force due to the quadriceps is 1795 N

<h3>What would happen if the vastus medialis was trained and strengthened to contract with 720N of force?</h3>

From the new information provided:

  • F₁ = 680N, θ = 90 = 30 = 120°
  • F₂ = 220 N, θ = 90 + 16 = 106°
  • F₃ = 600 N, θ = 90 + 15 = 105°
  • F₄ = 720 N, θ = 90 - 35 = 55°

then:

Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 720 * cos 55

Fx = -142.95 N

Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 720 * sin 55

Fₙ = 1969.72 N

then:

F = √(-142.95)² + ( 1969.72)²

F = 1974.9 N

F ≈ 1975 N

Therefore, the resultant force due to the quadriceps is 1975 N.

Training and strengthening the vastus medialis results in a greater force of muscle contraction.

Learn more about resultant of forces at: brainly.com/question/25239010

3 0
2 years ago
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