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elixir [45]
3 years ago
9

A pendulum of unknown mass is attached to the ceiling and nearly touches the floor as it completes 6 full cycles in 45 s. At its

lowest point, the pendulum moves at a speed of 4.15 m/s.
part a. What is the height of the ceiling?
part b. What is the maximum height of the pendulum?
part c. If the pendulum were moved to the moon, with a gravitational acceleration of 1.62 m/s2, what would be the resulting frequency?
part d. If the pendulum, back on Earth, were raised to a height of 0.22 m and released, what would be the resulting frequency of the motion?
Physics
1 answer:
lys-0071 [83]3 years ago
6 0

Answer:

a) h = 14 m

b) h = 88 cm

c) f = 0.054 Hz

d) f = 0.13 Hz

Explanation:

a) T = 2π√(L/g)

L = T²g/4π²

L = (45/6)²(9.8) / 4π² = 13.963...

b) ½mv² = mgh

h = v²/2g

h = 4.15²/ (2(9.8)) = 0.87869

c) f = 1/T = 1 / (2π√(14 / 1.62)) = 0.0542

d) f = 6/45 = 0.13333...

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Three types of spacesuits exist for different purposes: IVA (intravehicular activity), EVA (extravehicular activity), and IEVA (intra/extravehicular activity). IVA suits are meant to be worn inside a pressurized spacecraft, and are therefore lighter and more comfortable. IEVA suits are meant for use inside and outside the spacecraft, such as the Gemini G4C suit. They include more protection from the harsh conditions of space, such as protection from micrometeorites and extreme temperature change. EVA suits, such as the EMU, are used outside spacecraft, for either planetary exploration or spacewalks. They must protect the wearer against all conditions of space, as well as provide mobility and functionality.
8 0
3 years ago
Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli
Pie

Answer:

a)  v = 3,843 m / s, b)  46.7º  North- East

Explanation:

Moment is a vector quantity, so one of the best ways to solve this problem is to solve each component separately.

The system is formed by the two vehicles so that the moment is preserved during the crash

Direction to the East    

initial instant. Before the crash

          p₀ = mₐ vₐ₀

final insttne. After the crash

          p_f = (mₐ + m_b) vₓ

         p₀ = p_f

         mₐ vₐ₀ = (mₐ + m_b) vₓ

         vₓ = \frac{m_a}{m_a + m_b} \ v_{ao}

let's calculate

          vₓ = \frac{16.7}{16.7 + 29.3} \ 7.26

          vₓ = 2,636 m / s

direction north

initial   p₀ = m_b v_{bo}

final     p_f = (mₐ + m_b) v_y

          p₀ = p_f

          m_b v_{bo} = (mₐ + m_b) v_y

          v_y = \frac{m_b}{m_a+m_b} \ v_{bo}

let's calculate

          v_y = \frac{29.3}{16.7 + 29.3} \ 4.39

          v_y = 2.796 m / s

the final speed of the two two vehicles is

          v = (2,636 i ^ + 2,796 j ^) m / s

a) the magnitude of the velocity

let's use the Pythagorean theorem

       v = \sqrt{v_x^2 + v_y^2}

      v = \sqrt{2.636^2 + 2.796^2}

      v = 3,843 m / s

b) let's use trigonometry to find the direction

      tan θ = v_y / vₓ

      θ = tan⁻¹ v_y / vₓ

      θ = tan⁻¹ (2,796 / 2,636)

      θ = 46.7º

This direction is 46.7º  North East

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