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Dimas [21]
3 years ago
14

A rod of mass M = 4 kg, length L = 1.8 meters, and moment of inertia ML2/12 is free to move on a frictionless surface. The rod i

s at rest when a puck of mass m = 0.4 kg approaches with a speed vi = 20 m/s perpendicular to the rod's length and strikes the rod at a point d = 0.3 meters from its lower end. After the collision, the puck moves backward with a speed vf = 10 m/s. The rod's center of mass moves forward with speed v and rotates with angular speed ω. (a) What is the speed v of the rod's center of mass after the collision?
Physics
1 answer:
Pachacha [2.7K]3 years ago
8 0

Answer:

The speed of the rod's center of mass after the collision is 6 m/s.

Explanation:

Given that,

Mass of rod = 4 kg

Length l = 1.8 m

Moment of inertia I=\dfrac{ML^2}{12}

Mass of puck = 0.4 kg

Initial speed= 20 m/s

Distance = 0.3 m

Final speed = 10 m/s

(a). We need to calculate the speed v of the rod's center of mass after the collision

As there is no external force acting on the system so, linear and angular momentum of the system will be conserved.

Using conservation of momentum

m_{i}v_{i}=m_{f}v_{f}+Mv

Put the value into the formula

0.4\times20=-0.4\times10+2v

v=\dfrac{8+4}{2}

v=6\ m/s

Hence, The speed of the rod's center of mass after the collision is 6 m/s.

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A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the s
leonid [27]

Complete question is;

A force stretches a wire by 0.60 mm. A second wire of the same material has the same cross section and twice the length.

a) How far will it be stretched by the same force?

b) A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force?

Answer:

0.15 mm

Explanation:

According to Hooke's Law,

E = Stress(σ)/Strain(ε)

Where E is youngs modulus

Formula for stress is;

Stress(σ) = Force(F)/Area(A)

Formula for strain is;

Strain(ε) = Change in length/original length = (Lf - Li)/Li

We are also told that a second wire of the same material has the same cross section and twice the length.

Thus;

Rearranging Hooke's Law to get the constants on one side, we have;

F/(AE) = ε

Thus from the conditions given;

ε1 = 0.6/Li

ε2 = (Change in length)/(2*Li)

And ε1 = ε2

Thus;

0.6/Li = Change in length/(2*Li)

Li will cancel out and we now have;

Change in length = 2 × 0.6 = 1.2 mm

Finally, we are told A third wire of the same material has the same length and twice the diameter as the first.

Area of a circle;A1 = πd²/4

Now, we are told d is doubled.

Thus, new area of the new circle is;

A2 = π(2d)²/4 = πd²

Rearranging Hooke's Law,we have;

F/A = εE

Since F and E are now constants, we have;

F/E = constant = Aε

Thus;

A1(ε1) = A2(ε2)

A1 = πd²/4

e1 = 0.60/Li

A2 = πd²

e2 = Change in length/Li

Thus;

((πd²/4) × 0.6)/Li = (πd² × Change in length)/ Li

Rearranging, Li and πd² will cancel out to give;

0.6/4 = Change in length

Change in length = 0.15 mm

4 0
2 years ago
How many grams are in 6.53 moles of Mn?
Westkost [7]

Answer:

359 g Mn

General Formulas and Concepts:

  • Dimensional Analysis
  • Reading the Periodic Table of Elements

Explanation:

<u>Step 1: Define</u>

6.53 mol Mn

<u>Step 2: Find conversion</u>

1 mol Mn = 54.94 g Mn

<u>Step 3: Dimensional Analysis</u>

<u />6.53 \hspace{3} mol \hspace{3} Mn(\frac{54.94 \hspace{3} g \hspace{3} Mn}{1 \hspace{3} mol \hspace{3} Mn} ) = 358.758 g Mn

<u>Step 4: Simplify</u>

<em>We are given 3 sig figs.</em>

358.758 g Mn ≈ 359 g Mn

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What happens when the dew point and the temperature are the same?
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Answer:    

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<em>Impulse = Impulsive force *time</em>

I = F*Δt

If the object should be bought to rest from certain velocity there should be change in momentum. If the duration in which the momentum is increased then there would be less force applied and hence less damage.

Airbags are used to reduce the force experience by the people when they are met with accident by extending the time required to stop the momentum.

During the collision, the passenger is carried towards the<em> windshield</em> and if they are stopped by collision with wind shield the force will be larger and more damage.But if they are hit with airbag then the force will be less due to increased time.

The change is momentum will be the same with or without momentum but its the time that decides the impact of force.By making it longer the force become less.

<em>Thus compared to the windshield the airbag exerts much lesser force.</em>

<em>   </em>

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