Ali is whirling a 2.0 kg bunch of bananas in a circular path having a radius of 0.50 m. The bananas complete 2 revolutions every 6.0 seconds.
1. What force must he apply to keep the motion constant so that the bananas complete one revolution every 4 seconds?
2. A coin is on a turntable that rotates pi rad/s. The coefficient of static friction between the coin and the turntable is 0.25.
A. Calculation the maximum distance of the coin from the center of the turntable for it to move in a circle.
B. If the coin is placed at a distance of 4.7cm from the center of the turntable, what is the maximum speed of rotation of the turntable for the coin to move relative to the turntable without slipping?

Answers

Answer 1

Answer:

1) 2.467 N

2) a) 0.248m

   b) 2.3π rad/sec

Explanation:

Given data:

mass of Banana bunch ( m ) = 2.0 kg

radius of circular path ( R ) = 0.5 m

number of revolutions completed = 2

Time to complete 2 revolutions = 6 seconds

1) Determine the force to keep the motion constant for one complete revolution in every 4 seconds

F = mv^2 / r ----- ( 1 )

where V = 2πR/T

where : R = 0.5 , m = 2, T = 4 seconds

Insert values into equation 1

F = 2 * 4π^2 * 0.5/4^2

 = 2.467 N

2a) Calculate the maximum distance of coin from center

angular velocity ( w ) = v/r

coefficient of static friction  ( μ ) = 0.25

[tex]F_{c} = u mg[/tex]  ---- ( 1 )

mv^2/r = μmg --- ( 2 )        cancelling the mass on both sides eqn 2 becomes

v^2 = μ*g*r

dividing both sides of equation by r^2

w^2 = μ*g/r

hence determine distance ( r ) of coin from center

r = 0.25 * 9.81 / π^2 =  0.248 m

2b ) determine the maximum speed of rotation of the turntable for the coin to move relative to the turntable without slipping

distance coin is placed ( r ) = 4.7 cm = 0.047 m

find speed of rotation ( w )

w^2 =  μ*g/r

w = √ 0.25 * 9.81/ 0.047

   = 7.2236 rad/secs ≈  2.3π rad/sec

       


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During a fireworks display, a shell is shot into the air at an angle of theta degrees above the horizontal. The fuse is timed to 8.6 seconds for the shell just as it reaches its highest point above the ground. If the horizontal displacement of the shell is 81.7 meters when it explodes, what is the angle theta of the initial velocity

Answers

Answer:

θ = 83.6º

Explanation:

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       [tex]\frac{v_{o}* sin \theta}{v_{o}* cos \theta} = tg \theta= \frac{84.3m/s}{9.5m/s} = 8.87 (5)[/tex]

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Answer:

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Answer:

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Explanation:

Answer:

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Answer:

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Answer:

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Answers

Answer:

A. nuclear fusion reactions

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Answers

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1.
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Answers

Answer:

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Explanation:

A 5 kilogram block is dropped from a height of 3 meters and falls straight to the ground. What is the work done by the force of gravity?

Answers

Answer:

Workdone = 147Nm

Explanation:

Given the following data;

Mass = 5kg

Height = 3m

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[tex] Force = mass * acceleration [/tex]

Force = 5*9.8

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Now, to find the workdone;

[tex] Workdone = force * distance [/tex]

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Answers

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Answers

Question:

A small rocket is launched vertically upward from the edge of a cliff 80 ft off the ground at a speed of 69 ft/s. Its height (in feet) above the ground is given by  [tex]h(t) = - 25t^2 + 200t + 400[/tex] where t represents time measured in seconds.

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[tex]0 \le t \le 9.66[/tex]

Explanation:

Given

[tex]h(t) = - 25t^2 + 200t + 400[/tex]

Required

Determine the domain of h

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Using quadratic formula:

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The expression becomes:

[tex]t = \frac{-(-8)\±\sqrt{(-8)^2 - 4*1*(-16)}}{2*1}[/tex]

[tex]t = \frac{8\±\sqrt{64+ 64}}{2}[/tex]

[tex]t = \frac{8\±\sqrt{128}}{2}[/tex]

[tex]\sqrt{128} = 11.31[/tex]

So, we have:

[tex]t = \frac{8\±11.31}{2}[/tex]

Split

[tex]t = \frac{8+11.31}{2}\ or\ t = \frac{8-11.31}{2}[/tex]

[tex]t = \frac{19.31}{2}\ or\ t = \frac{-3.31}{2}[/tex]

[tex]t = 9.66\ or\ t = -1.66[/tex]

But time can't be negative.

So:

[tex]t = 9.66[/tex]

Hence, the domain is:

[tex]0 \le t \le 9.66[/tex]

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