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inna [77]
3 years ago
6

A block with a mass of 0.400 kg is connected to a spring, displaced in the positive direction a distance of 50.0 cm from equilib

rium, and released from rest at t = 0. The block then oscillates without friction on a horizontal surface. After being released, the first time the block is a distance of 25.0 cm from equilibrium is at t = 0.200 s.
(a) What is the block's period of oscillation?

(b) What is the the value of the spring constant?
Physics
1 answer:
Gemiola [76]3 years ago
7 0

Answer:

a)  T = 1,199 s b)  k = 11.0 N / m

Explanation:

The oscillatory movement of a mass-spring system is described by the equation

    x (t) = A cos (w t)

Where A is the amplitude of the movement. W is the angular velocity, which is given by

    w = √ k / m

The amplitude of the system is the maximum elongation that the spring has before releasing it, in this case it is A = 50 cm = 0.50 m, the equation is

     x = 0.50 cos (wt)

We can calculate the angular velocity with the point given by x = 0.250 m for t = 0.200s

     w t = cos⁻¹ (x / A)

     w = 1 / t cos⁻¹ (x / A)

     w = 1/0.200  cos⁻¹(0.25 / 0.50)

Let's be careful because the angle is in radians

     w = 5.24 rad / s

Angular velocity is

    w = 2 π f = 2π / T

    T = 2π / w

    T = 2π / 5.24

    T = 1,199 s

With the angular velocity equation we can take off the spring constant

    w² = k / m

    k = m w²

    k = 0.400 5.24²

    k = 11.0 N / m

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What is a stable and unstable equilibrium
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Answer:

here

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A four-wheel-drive vehicle is transporting an injured hiker to the hospital from a point that is 30 km from the nearest point on
zepelin [54]

Answer:

D=200\ km

Explanation:

distance on terrain, d_t=30\ km

  • distance on the road, d_r=70\ km
  • speed on terrain, v_t=30\ km.hr^{-1}
  • speed  on road, v_r=130\ km.hr^{-1}

<u>time taken on the terrain,</u>

t_t=\frac{d_t}{v_t}

t_t=\frac{30}{30}

t_t=1\ hr

<u>time taken to cover the distance on the road:</u>

t_r=\frac{d_r}{v_r}

t_r=\frac{70}{130}

t_r=\frac{7}{13}\ hr

<u>Now the distance covered on terrain in the total time:</u>

D= v_r\times (t_r+t_t)

D= 130\times (\frac{7}{13}+1)

D=130\times \frac{20}{13}\

D=200\ km

<em>is the distance the vehicle must target on the road to minimize the time taken in going off the road.</em>

3 0
3 years ago
A rocket sled with an initial mass of 3 metric tons, invluding 1 ton of fuel, rests on a level section of track. At t=0, the sol
Katena32 [7]

Answer:

v = 719.2 m / s and     a = 83.33 m / s²

Explanation:

This is a rocket propulsion system where the system is made up of the rocket plus the ejected mass, where the final velocity is

           v - v₀ = v_{e} ln (M₀ / M)

where v₀ is the initial velocity, v_{e} the velocity of the gases with respect to the rocket and M₀ and M the initial and final masses of the rocket

In this case, if fuel burns at 75 kg / s, we can calculate the fuel burned for the 10 s

            m_fuel = 75 10

            m_fuel = 750 kg

As the rocket initially had a mass of 3000 kg including 1000 kg of fuel, there are still 250 kg, so the mass of the rocket minus the fuel burned is

              M = 3000 -750 = 2250 kg

let's calculate

            v - 0 = 2500 ln (3000/2250)

            v = 719.2 m / s

To calculate the acceleration, let's use the concept of the rocket thrust, which is the force of the gases on it. In the case of the rocket, it is

             Push = v_{e} dM / dt

let's calculate

             Push = 2500  75

             Push = 187500 N

            If we use Newton's second law

             F = m a

             a = F / m

let's calculate

              a = 187500/2250

              a = 83.33 m / s²

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valentinak56 [21]
Try chegg they have answers
6 0
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