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Allisa [31]
3 years ago
8

A 2.0 kg bucket is attached to a horizontal ideal spring and rests on frictionless ice. You have a 1.0 kg mass

Physics
1 answer:
bogdanovich [222]3 years ago
8 0

Answer:

x = A cos (w \sqrt{2y_{o}/g})

a) maximun  Ф= \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) minimun     Ф = \frac{\pi }{2} - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

Explanation:

For this exercise let's use kinematics to find the time it takes for the mass to reach the floor

         y = y₀ + v₀ t - ½ g t²

   

as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)

      0 = y₀ - ½ g t²

      t = \sqrt{2y_{o}/g}

The bucket-spring system has a simple harmonic motion, which is described by

     x = A cos wt

in this expression we assumed that the phase constant (Ф) is zero

let's replace the time

     x = A cos (w \sqrt{2y_{o}/g})

this is the distance where the system must be for the mass to fall into it.

a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes

          w = \sqrt{k/m}

In the initial state

         w = \sqrt{k/2}

When the mass changes

         w ’= \sqrt{k/3}

the displacement in each case is

         x = A cos (wt)

for the new case

        x ’= A cos (w’t + Ф)

the phase constant is included to take into account possible changes due to the collision of the mass.

we see that this maximum expressions when the cosine is maximum

        cos (w´t + Ф) = 1

         w’t + Ф = 0

        Ф = -w ’t

        Ф = - \sqrt{k/3} \sqrt{2y_{o}/g}

       \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) the function is minimun if

        cos (w’t + fi) = 0

        w’t + Ф = π / 2

        Ф = π / 2 - w ’t

        Ф = \frac{\pi }{2} - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

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satela [25.4K]

Answer:

Changes in the object's momentum (answer D)

Explanation:

A net force will cause an object to change its velocity, and that will affect the object's momentum, which is defined by the product of the object's mass times its velocity.

So, select the last option (D) in the given list.

8 0
3 years ago
A ship sets out to sail to a point 120 km due north. an unexpected storm blows the ship to a point 100 km due east of its starti
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If you draw the problem, it would look like that shown in the attached picture. The total length the ship will now travel can be solved using the Pythagorean theorem. The solution is as follows:

d = √(120)²+(100)²
d = 156.2 km

So, the ship will have to travel 156.2 km to the northwest direction.

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3 years ago
The spool has a mass of 50 kg and a radius of gyration of ko = 0.280 m. if the 20-kg block a is released from rest, determine th
Rzqust [24]
V₁ = (1/g)₁ = Way₁ = 20(9.81)(0) = 0
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The kinetic energy because the pool rotates about a fixed axis 
W = VA/rA = VA/0.2 5VA
Mass momen of inertila about fixed axis which passes through point 0
I₀ = mko² = 50(0.280)² = 3.92 kg. m²
∴ The kinetic energy of the system is 
T = 1/2 I₀w² + 1/2mAVA²

= 1/2(3.92)(5VA)² + 1/2 (20) VA² = 59VA²
Now that the system is at rest then T₁ = 0
Energy conservation  is
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5 0
3 years ago
A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. If
ella [17]

Answer:

442.5 rad

Explanation:

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α = Constant angular acceleration = 3.0 rad/s²

t = time period of rotation of the disk = 15 s

θ = angular displacement of the point on the rim

Angular displacement of the point on the rim is given as

θ = w₀ t + (0.5) α t²

inserting the values

θ = (7.0) (15) + (0.5) (3.0) (15)²

θ = 442.5 rad

4 0
3 years ago
A copper wire has a radius of 3.5 mm. When forces of a certain equal magnitude but opposite directions are applied to the ends o
denpristay [2]

Answer:

The tensile stress on the wire is 550 MPa.

Explanation:

Given;

Radius of copper wire, R = 3.5 mm

extension of the copper wire, e =  5.0×10⁻³ L

L is the original length of the copper wire,

Young's modulus for copper, Y =  11×10¹⁰Pa.

Young's modulus, Y is given as the ratio of tensile stress to tensile strain, measured in the same unit as Young's modulus.

Y =\frac{Tensile \ stress}{Tensile \ strain} \\\\Tensile \ stress = Y*Tensile \ strain\\\\But, Tensile \ strain = \frac{extension}{original \ Length} = \frac{5.0*10^{-3} L}{L} = 5.0*10^{-3}\\\\Tensile \ stress = 11*10^{10} *5.0*10^{-3} \ = 550*10^6 \ Pa

Therefore, the tensile stress on the wire is 550 MPa.

8 0
3 years ago
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