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3 years ago

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A 24-foot simple span laterally supported beam is required to carry a uniformly distributed load of 8.0 klf. Select the lightest

A992 steel beams for each of these combinations of load considering only flexure. 1 klf 3 klf 5 klf 7 klf DL 7 klf 5 klf 3 klf 1 klf C.

1 answer:

vampirchik [111]3 years ago

8 0

Answer:

1 klf

Explanation:

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<h3>Yes</h3>

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If you build thing "a" and thing "a" builds thing "b" you <u>indirectly</u> build thing "b".

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3 years ago

A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust end

hjlf

Answer:

a) $v=19.6 m/s$

b) $H=19.58 m$

c) $v_{f}=29.57 m/s$

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

$W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J$

But we know the work is equal to change of kinetic energy, so:

$W=\Delta K=\frac{1}{2}mv^{2}$

$v=\sqrt{\frac{2W}{m}}=19.6 m/s$

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

$v_{f}^{2}=v_{i}^{2}-2gh$

At the maximum height the velocity is 0, so v(f) = 0.

$0=v_{i}^{2}-2gH$

$H=\frac{19.6^{2}}{2*9.81}=19.58 m$

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

$v_{f}^{2}=v_{i}^{2}-2gh$

$v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s$

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

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3 years ago