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natka813 [3]
3 years ago
5

Here on Earth you hang a mass from a vertical spring and start it oscillating with amplitude 1.9 cm. You observe that it takes 3

.0 s to make one round trip. You construct another vertical oscillator with a mass 6 times as heavy and a spring 10 times as stiff. You take it to a planet where gplanet = 8.8 N/kg. You start it oscillating with amplitude 4.8 cm. How long does it take for the mass to make one round trip?
Physics
1 answer:
Vlada [557]3 years ago
5 0

Answer:

T = 3.23 s

Explanation:

In the simple harmonic movement of a spring with a mass the angular velocity is given by

               w = √ K / m

With the initial data let's look for the ratio k / m

The angular velocity is related to the frequency and period

           w = 2π f = 2π / T

            2π / T = √ k / m

            k₀ / m₀ = (2π / T)²

            k₀ / m₀ = (2π / 3.0)²

            k₀ / m₀ = 4.3865

The period on the new planet is

          2π / T = √ k / m

           T = 2π √ m / k

In this case the amounts are

           m = 6 m₀

           k = 10 k₀

We replace

            T = 2π√6m₀ / 10k₀

            T = 2π √6/10 √m₀ / k₀

            T = 2π √ 0.6  √1 / 4.3865

            T = 3.23 s

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If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is u  =  7.73 \  m/s

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      s = h+  u t +  0.5 gt^2

Here s  is the distance covered by the bag at sea level which is zero

      0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>    0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>   u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}

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A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
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Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

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now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

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from equation (1)

T₂ = 1176 - 710.5

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hence, T₁ = 710.5 N and T₂ = 465.5 N

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