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ioda
3 years ago
3

An object initially at rest falls from a height H until it reaches the ground. Two of the following energy bar charts represent

the kinetic energy K and gravitational potential energy Ug of the object-Earth system at two positions. The first position is when the object is initially released, and the second position is when the object is halfway between its release point and the ground. Which two charts could represent the mechanical energy of the object-Earth system? Select two answers. by Ug K Ug к Mechanical Energy at Halfway Point Mechanical Energy at Release Point Mechanical Energy at Release Point Mechanical Energy at Halfway Point d. c. к Ug Ug K Mechanical Energy at Release Point Mechanical Energy at Halfway Point Mechanical Energy at Release Point Mechanical Energy at Halfway Point
Physics
1 answer:
Tom [10]3 years ago
6 0

Answer:

Initial:   bar  power U₀

Final:    bar  power U = U₀ / 2

             bar Kinetic energy  K = U₀ / 2

These two bars are half the height of the initial bar

Explanation:

To know which graph is correct, let's discuss the solution to the problem

Initial mechanical energy

      Em₀ = U₀ = m g H

The mechanical energy at the midpoint

     Em₂ = K + U₂

As there is no friction, mechanical energy is conserved

     Em₀ = Em₂

     U₀ = K + U₂

     K = U₀ - U₂

     K = m g (H - y₂)

Indicates that position 2 corresponds to y₂ = H / 2

     K = m g (H –H / 2)

     K = ½ m g H

     K = ½ Uo

Therefore the graph must be

Initial:   bar  power U₀

Final:    bar  power U = U₀ / 2

             bar Kinetic energy  K = U₀ / 2

These two bars are half the height of the initial bar

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Answer:

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Explanation:

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for dark fringe,we have

y_m = {(m + 1/2)\lambda D}{d}    

Now to get the dark fringes at the same location we should have;

(706)D/d = (m + 1/2)λD/d    

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4 years ago
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Answer:

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Explanation:

Since F₁ = 9.2 N and acts at 57° above the negative axis in the second quadrant, its x-component is -F₁cos57° and its y- component is F₁sin57°

Since F₁ = 5.2 N and acts at 53.7° below the negative axis in the third quadrant, its x-component is -F₂cos53.7° and its y- component is -F₂sin53.7°

Part A

What is the x component Fx of the resultant force?

The x component of the resultant force Fx = -F₁cos57° + -F₂cos53.7° = -9.2cos57° + (-5.2cos53.7°) = (-5.011 - 3.078) N = -8.089 N

Part B

What is the y component Fy of the resultant force?

The y component Fy of the resultant force = F₁sin57° + -(F₂sin53.7°) = 9.2sin57° - 5.2sin53.7° = (7.716 - 4.191) N = 3.525 N

Part C  

What is the magnitude F of the resultant force?

The magnitude F of the resultant force = √(Fx² + Fy²)

F = √(-8.089² N + 3.525² N) = √65.432 + 12.426 = √77.858 = 8.824 N

Part D

What is the angle ? that the resultant force forms with the negative x axis?

The angle the resultant force makes with the negative x axis is given by

θ = tan⁻¹(Fy/Fx) = tan⁻¹(3.525/-8.089) = tan⁻¹-0.4358 = -23.55°.

To measure it from the negative x axis, we add 360. So, our angle = 360 -23.55 = 336.45°

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3 years ago
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3 years ago
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Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :

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