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irga5000 [103]
3 years ago
15

What were some of the challenges to safety resulting from such radical airframe designs as highly swept wings, high wing loading

, increasing speed, and long-duration flights are high altitude
Engineering
1 answer:
Andrew [12]3 years ago
8 0
How to take off and land, stopping considerations (stopping distance), control system capability over a large speed range and flutter, and structural integrity for the wing platform and speed range.
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Anthony and Angel are making plans for Friday. Anthony wants to go bowling. Angel doesn't bowl well and prefers to go ice-skatin
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I am mostly on the side of compromise do i think that’s the answer
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16. Which polymers are recyclable?
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Number 2 is HDPE, or High-Density Polyethylene. It is most commonly found in plastic bags, oil and detergent bottles, and toys. It is one of the simplest polymers to recycle. It is broken down into flakes, decontaminated and then reused for pipes, flower pots or reused as (non-food) bottles.

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3 years ago
A Carnot power cycle is executed in a piston-cylinder system using 1 kg of water. The cycle consists of an isobaric expansion of
jeka94

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b) Determine the heat transfer into the cycle and the net work for the cycle, in kJ.

Explanation:

5 0
4 years ago
A method that uses low temperature heat-treating that imparts toughness without reduction in hardness is called:_______
lbvjy [14]

Answer:

"Tempering Process" seems to be the appropriate choice.

Explanation:

  • Tempering seems to be a method of heat preparation which is mostly used in completely hard materials to increase consistency, strength, durability, and also some decreasing brittleness.
  • The tempering method is used to examine good functionality as well as flexural by reducing stiffness again after the substance has indeed been quenched towards its toughest state.
5 0
3 years ago
The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta
tatyana61 [14]

Answer:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular  momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

H_o =r x mv=rxL

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change  of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular  momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2".

If we analyze the staritning point we see that the initial velocity can be founded like this:

v_o =\omega r_{OIC}=\omega (0.15m)

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af

0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)

And if we integrate the left part and we simplify the right part we have

1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega

And if we solve for \omega we got:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

8 0
3 years ago
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