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3 years ago

6

A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o

f elasticity is 200 GPa, what is the change in length?

1 answer:

Crank3 years ago

7 0

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

$\delta = \frac{PL}{AE}$

Where,

P = Axial Load

l = Gage length

A = Cross-sectional Area

E = Modulus of Elasticity

Our values are given as,

l = 3.5m

D = 0.028m

$P = 68*10^3 N$

E = 200GPa

$A = \frac{\pi}{4}(0.028)^2 \rightarrow 0.0006157m^2$

Replacing we have,

$\delta = \frac{PL}{AE}$

$\delta = \frac{( 68*10^3)(3.5)}{(0.0006157)(200*10^9)}$

$\delta = 0.001932m$

$\delta = 1.93mm$

Therefore the change in length is 1.93mm

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3 years ago

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GaryK [48]

Answer:

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Explanation:

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3 years ago

Can someone answer these

LenKa [72]

Four

Sometimes I think the creators of problems out to drawn and quartered. 60 g does not mean 60 grams. It means 60 * the acceleration due to gravity.

So the question really reads. The acceleration delivered by the air bag is 60 times that of a normal gravitational. This acceleration is delivered to the person where his mass is putting up a whole lot of resistance because he and his 75 kg are moving forward with the impact of the car. The 36 msec. has nothing to do with the problem.

The Force of the Air Bag is mass * a

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The answer is 4.41 * 10^4

Answer C

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This problem is governed by one formula that you sort of have to get out of your hat -- a piece of magic if you will.

Fg - Bf = m * a

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Bf = the braking force

The mass of the rocket is derived from its weight

The acceleration is derived from one of your big 4 equations.

m of the rocket = 75600 / 9.81 = 7706 kg

The acceleration =

vi = 1 km/s = 1000 m/s

vf = 0

t = 2 minute * 60 sec/ min = 120 seconds

a = (vf - vi)/t = (0 - 1000 m/s) / 120 sec

a = - 8.333 m/s^2 The minus sign makes perfect sense. Remember the rocket is slowing down

The net downward force = mass * acceleration = - 7706 kg * - 8.333 m/s^2

The net force = - 64217 N

So going back to the problem's equation we have

Gravitational force - Braking Force = Net Force

Gravitational Force = 75600

Net Force = - 64217

Bracking force = ?

75600 - Bracking force = - 64217 Subtract 75600 from both sides

- Bracking force = - 64217 - 75600

- Braking force = - 139817

Braking force = 139817 N = 1.398 * 10^5 N

Braking Force = 1.4 * 10^5

Answer: Last One.

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The first thing you should do is derive a general formula for this problem.

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The formula for this problem is

Mg = (m + M) * a

Now you need to solve for a

a = [M/(M + m) ] * g

Look what is happening. is a smaller or larger than g? This is a question you should really pay attention to. If it was larger, everyone would have this system in their basement because you'd get more energy output than you put in. Something for nothing is always appealing.

So what's the answer? (I get to ask it. No one posing the question ever should).

A

A is incorrect. M never goes away. The acceleration may get very tiny, but there always is some acceleration.

B must be true. It is just what I finished saying about A

C Who said anything about velocity? It's a red herring. If the velocity became 0 the acceleration would have to turn minus. This answer sounds good, but sounds good doesn't make it right. C is wrong.

D The acceleration does not remain constant no matter what. The answer to A still applies. So D is wrong.

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2 years ago

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Katyanochek1 [597]

Answer:

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Explanation:

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2 years ago

An experiment is designed to compare the differences in learning outcomes

MissTica

Answer:

B. Randomly assigning students to the two different groups

Explanation:

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3 years ago