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Nady [450]
3 years ago
12

A bullet with a mass of 4.5 g is moving with a speed of 300 m/s (with respect to the ground) when it collides with a rod with a

mass of 3 kg and a length of L = 0.25 m. The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass which is located exactly halfway along the rod. The bullet imbeds itself in the rod at a distance L/4 from the pivot point. As a result, the bullet-rod system starts rotating together. What is the magnitude of the angular velocity (in rad/s) of the rotation (with respect to the ground) immediately after the collision? You may treat the bullet as a point particle.
Physics
1 answer:
Law Incorporation [45]3 years ago
5 0

Answer:

ω = 5.41 rad/s

Explanation:

Since the rod is rotating around its axis, angular momentum will play part in this question.

The conservation of angular momentum implies that

L_1 = L_2

So, the initial angular momentum is

L_1 = m_b v_b \frac{L}{4} = (4.5\times 10^{-3}~{\rm kg})(300~{\rm m/s})(\frac{0.25}{4}~{\rm m}) = 0.0844~{\rm kg.m^2/s}

The final angular momentum includes the rod and the bullet together. So,

L_2 = I\omega\\I = I_{rod} + I_{bullet} = \frac{1}{12}m_r L^2 + m_b(\frac{L}{4})^2 = \frac{1}{12}3(0.25)^2 + (4.5\times 10^{-3})(\frac{0.25}{4})^2 = 0.0156~{\rm kg.m^2}\\L_2  = L_1 = I\omega\\0.0844 = 0.0156\omega\\\omega = 5.41~{\rm rad/s}

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A cubical wooden box floating on water rises 1cm when 400 gm of stone is
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Answer:

Explanation:

I'm not  sure you can do this without just a bit more information. I can tell you what the mass of the water is when the rocks are removed. When we know that, we know the volume of the water that was displaced. whether or not this is enough information to determine the volume of the box, I'm not sure.

400 grams raises the box 1 cm.

The density of water = 1 gm / cm^3

400 grams of water = 400 mL or 400 cm^3

The volume of the displaced water = 400 cm^3

The volume a slice from the square prism is B*h

B = 400 cm^2

h = 1 cm

If the base is 400 cm^2 then each side is

s^2 = 400

sqrt(s^2)= sqrt(400)

s = 20

The volume of the box is 20^3 = 8000 cm^3

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3 years ago
A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
jeka57 [31]

Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

a.

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A²  ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

b.

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

c.

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J  = 120 J decrease

d.

The missing energy has turned into interned energy in the completely inelastic collision

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3 years ago
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ELEN [110]
<span>ultraviolet

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rewona [7]

Answer and Explanation:

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konstantin123 [22]

The work done on the puck is 96 J

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According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

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u = 2 m/s being the initial speed of the puck

Substituting numbers into the equation, we find the work done by the player on the puck:

W=\frac{1}{2}(2)(10)^2 - \frac{1}{2}(2)(2)^2=96 J

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