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mario62 [17]
3 years ago
6

1. Read each description below. Choose the force diagram (free-body diagram) that best represents the description. The forces ar

e acting on the object in red type. You may neglect the effects of air resistance.
A. Mindy's Christmas tree ornament hangs motionless on a tree.
B. Corey's Christmas tree ornament is falling to the floor.
C. A ball thrown by Ginger is moving upward through the air.
D. A ball dropped by Melissa is falling downward toward the floor.
E. Matt's book is motionless on a table.
F. A ball that was thrown upward by Yvonne is at the top of its path.

For each statement, please choose a diagram representing the force on the object. (Diagrams are in the attachment)

Physics
1 answer:
pishuonlain [190]3 years ago
7 0
<span>A. Mindy's Christmas tree ornament hangs motionless on a tree. --- Diagram C
B. Corey's Christmas tree ornament is falling to the floor. </span>--- Diagram A<span>
C. A ball thrown by Ginger is moving upward through the air. </span>--- Diagram B<span>
D. A ball dropped by Melissa is falling downward toward the floor. </span>--- Diagram A<span>
E. Matt's book is motionless on a table. </span>--- Diagram C<span>
F. A ball that was thrown upward by Yvonne is at the top of its path </span>--- Diagram A.
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The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

  • The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s
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<h3>How can the velocity and height of the block be calculated?</h3>

Mass of the block, m = 3 kg

Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m}  = \mathbf{ 1000\, N/m}

Coefficient of kinetic friction, \mu_k = 0.39

Therefore, we have;

Friction force = \mathbf{\mu_k}·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, <em>W</em> ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

The initial velocity of the block is therefore;

5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, <em>W</em> ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

  • The velocity of the block after the Spring extends 7 cm, v₂ ≈ <u>1.73 m/s</u>

The height at which the block will stop moving, <em>h</em>, is given as follows;

At \ the \ maximum \ height, \ h, \ we \ have ; \  \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x

Which gives;

Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{  7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2  } \approx 1.536

The distance up the inclined, the block rises, at maximum height is therefore;

x_{max} ≈ 1.536 m

Therefore;

h = 1.536 × sin(48°) ≈ 1.1415

  • The height at which the block stops rising, h ≈ <u>1.1415 m</u>

From the above solution for the height, the length of the incline is he

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Py= 5,30 * 10^{-3} * 301 = 1,5953 kg*m/s

The angle between x and y bullets is 120°, and we know that if the angle between two equal magnitude vectors is 120°, the magnitude of the resultant vector will be equal to first two and placed in exact middle of two vectors. So we can say total momentum of x and y (Px+Py) equals to 1,5953 kg*m/s as well (Shown in the figure).  

For z bullet to equalize the total momentum of x and y bullets, it needs to have the same amount of momentum in the opposite way.

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The amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal on the top and the bottom is 0.1 cm thick and the metal on the sides is 0.05 cm thick is 8.8 cm.

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V = πr^2 h

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V/ h  = 2  πr h

Now, the differential becomes

dV =  πr^2dh +  2πrh dr

Given the following parameters i.e. diameter and height

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dr = 0.05 cm

h = 10 cm

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r = 2cm

Substituting the values in the above equation, we get

dV = 3.14(2)^2(0.2)  + 2(3.14)(2)(10)(0.05)

dV = 2.512 + 6.28

dV = 8.792 cm

dV = 8.8 cm

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