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inn [45]
3 years ago
8

A glider with mass m = 0.230 kg sits on a frictionless horizontal air track, connected to a spring with force constant k = 4.50

N/m . You pull on the glider, stretching the spring 0.130 m , and then release it with no initial velocity. The glider begins to move back toward its equilibrium position (x=0). What is the speed of the glider when it returns to x=0?
What must the initial displacement of the glider be if its maximum speed in the subsequent motion is to be 3.00 m/s?
Physics
1 answer:
loris [4]3 years ago
8 0

Answer

given,

mass of glider = 0.23 Kg

spring constant = k = 4.50 N/m

spring stretched to 0.130 m

The springs potential energy =

 U = \dfrac{1}{2}kx^2

 U = \dfrac{1}{2}\times 4.5 \times 0.13^2

        U = 0.038 J

at x = 0,the only energy will be kinetic .

 \dfrac{1}{2}mv^2=0.038

 \dfrac{1}{2}\times 0.23 \times v^2=0.038

         v² = 0.3304

         v = 0.575 m/s

displacement of the glider

      using conservation of energy

 \dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2

 x =v\sqrt{\dfrac{m}{k}}

 x =3\times \sqrt{\dfrac{0.23}{4.5}}

        x = 0.678 m

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An unfortunate astronaut loses his grip during a spacewalk and finds himself floating away from the space station, carrying only
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Answer:

15.75 m/s

Explanation:

v = Velocity of the combined mass of astronaut and tools = 1.8 m/s

m_1 = Mass of astronaut = 124 kg

m_2 = Mass of tools = 16 kg

v_1 = Velocity of astronaut = 0

v_2 = Velocity of tools

As linear momentum is conserved

m_1v_1 + m_2v_2 =(m_1 + m_2)v\\\Rightarrow v_2=\frac{(m_1 + m_2)v-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{(124+16)\times 1.8-124\times 0}{16}\\\Rightarrow v_2=15.75\ m/s

The velocity of the tools is 15.75 m/s

3 0
3 years ago
A box of mass 12 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.42. What is th
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The maximum static force that can be applied is equal to the normal force*the frictional force. the normal force on the box is equal to mg since the floor is flat using 9.81m/s^2 for gravity 12kg*9.81m/s^2 = 118N multiplying the normal force by the frictional force you get a 118*.42= 49.6N so overcome the force of static friction on the box a minimum of 49.6N would need to be applied.
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A 7000-kg plane is launched from an aircraft carrier in 2.0 seconds
ExtremeBDS [4]

The acceleration and velocity of the plane is 78.57 m/s² and 157.14 m/s respectively

To calculate the acceleration of the plane, we use the formula below.

<h3>Formula:</h3>
  • a = F/m..................... Equation 1

Where:

  • a = Acceleration of the plane
  • F = Force applied to the plane
  • m = mass of the plane.

From the question,

Given:

  • F = 550000 N
  • m = 7000 kg

Substitute these values into equation 1

  • a = 550000/7000
  • a = 78.57 m/s²

To calculate the velocity, we use the formula below.

  • v = u+at............. Equation 2

Where:

  • v = Final velocity
  • u = initial velocity
  • a = acceleration
  • t = time.

From the question,

Given:

  • u = 0 m/s
  • a = 78.57 m/s
  • t = 2.0 seconds

Substitute these values into equation 2

  • v = 0+2(78.57)
  • v = 157.14 m/s

Hence, The acceleration and velocity of the plane is 78.57 m/s² and 157.14 m/s respectively.
Learn more about acceleration here: brainly.com/question/460763

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