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Paul [167]
3 years ago
12

A uniform thin rod of length 0.60 m and mass 3.5 kg can rotate in a horizontal plane about a vertical axis through its center. T

he rod is at rest when a 3.0 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 14 rad/s immediately after the collision, what is the bullet's speed just before impact? (Answer is in m/s)
Physics
1 answer:
abruzzese [7]3 years ago
6 0
 <span>The moment of inertia for the rod is 
mr*L^2/12 
After collision 
I=mr*l^2/12+mb*r^2 
I=0.073 
The momentum of the rod after collision is 
I*w 
=1.022 

The momentum of the bullet that is radial 
to the rod is 
vr=v*sin(60) 
vr/r=w 

so the radial momentum of the bullet is 
3*vr/(0.25*1000) 

vr*0.012 

applying conservation of momentum 
vr*0.012=1.022 
vr=85 m/s 

and 
v=85/sin(60) 
v=98 m,/s 

j</span>
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kow [346]

Answer:

Conservation of angular momentum

Explanation:

When the objects spread in universe after big bang, because of the tremendous force , they gained angular momentum and started to rotate. Since, then the object continue to rotate on their axis because of conservation of angular momentum. In vacuum of space there no other forces that can stop these rotation, therefore, they continue to rotate.  

4 0
3 years ago
A mass of 8 kg moves at a rate of 14.3 m/sec. what is the KE developed by the mass?
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3 years ago
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an
nalin [4]

Answer:

The maximum volume is 1417.87 inch^3

Explanation:

<u>Optimization Using Derivatives</u>

We have a 24x30 inch piece of metal and we need to make a rectangular box by cutting a square from each corner of the piece and bending up the sides. The width of the piece is 24 inches and its length is 30 inches

When we cut a square of each corner of side x, the base of the box (after bending up the sides) will be (24-2x) and (30-2x), width and length respectively. The volume of the box is

V=(24-2x)(30-2x)x

Operating

V=4x^3-108x^2+720x

To find the maximum value of V, we compute the first derivative and equate it to zero

V'=12x^2-216x+720=0

Simplifying by 12

x^2-18x+60=0

Completing squares

x^2-18x+81-81+60=0

(x-9)^2=21

We have two values for x

x=9+\sqrt{21}=13.58\ inch

x=9-\sqrt{21}=4.42\ inch

The first value is not feasible because it will produce a negative width (24-2(13.58))=-6.16

We'll keep only the solution

x=4.42\ inch

The width is

w=(24-2(4.42))=15.16\ inch

The length is

l=(30-2(4.42))=21.16\ inch

And the height

x=4.42\ inch

The maximum volume is

V=(15.16)(21.16)(4.42)=1417.87\ inch^3

4 0
3 years ago
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kvv77 [185]

Answer:

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Explanation:

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