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3 years ago

8

An aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and i

s 100mm long. If the modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180kN.

1 answer:

larisa86 [58]3 years ago

8 0

Answer: 1.228mm

Explanation:

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To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

$R = 350ft$

$a_t = 1.1ft/s^2$

PART A )

$a_c = \frac{V^2}{R}$

$a_c = \frac{V^2}{350}$

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is $5.25ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$27.5625 = 1.21 + \frac{v^4}{122500}$

$v=42.3877ft/s$

Now calculate the angular velocity of the motorcycle

$v = r\omega$

$42.3877 = 350\omega$

$\omega = 0.1211rad/s$

Calculate the angular acceleration of the motorcycle

$a_t = r\alpha$

$1.1 = 350\alpha$

$\alpha = 3.1428*10^{-3}rad/s^2$

Calculate the time needed by the motorcycle to reach an acceleration of

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$\omega = \alpha t$

$0.1211 = 3.1428*10^{-3}t$

$t = 38.53s$

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is $6.75ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$45.5625 = 1.21 + \frac{v^4}{122500}$

$v=48.2796ft/s$

PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is $21.5ft/s$

$a_r = \frac{v^2}{R}$

$a_r = \frac{21.5^2}{350}$

$a_r =1.3207ft/s^2$

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is $21.5ft/s$

$a = \sqrt{a_t^2+a_r^2}$

$a = \sqrt{(1.1)^2+(1.3207)^2}$

$a = 1.7187ft/s^2$

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is $6.75ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$45.5625 = 1.21 + \frac{v^4}{122500}$

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3 years ago

Is the average speed of a vehicles vector or a scalar quantity? A) vectorB) scalar

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As the speed is a scalar quantity as it has the only magnitude in it. Therefore, the average speed is also stated as a scalar quantity.

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1 year ago

A car traveling on a straight road at 15.0 meters per second accelerates uniformly to a speed of 21.0 meters per second in 12.0

oksano4ka [1.4K]

<h2>Option 3, 216 m is the correct answer.</h2>

Explanation:

We have initial velocity, u = 15 m/s

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Substituting

21 = 15 + a x 12

a = 0.5 m/s²

Now we have equation of motion v² = u² + 2as

21² = 15² + 2 x 0.5 x s

s = 216 m

Displacement = 216 m

Option 3, 216 m is the correct answer.

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4 years ago

A box experiencing a gravitational force of 600 N is being pulled to the right with force of 250 N. A 25 N frictional force acts

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The answer is: 0 Newtons

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Kipish [7]

Answer:

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The 6 will be positive, and the 10 and 2 will be negative.

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3 years ago

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