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Topic: **Wind homework****Question:**
A jet airline moving initially at 300 mi/h due east enters a region where the wind is blowing at 100 mi/h in a direction 30.0 degrees north of east. What is the new velocity of the aircraft relative to the ground?

June 17, 2019 / By Darrin

Sounds like homework ! I suggest you look up vector mathematics via text book or the internet and see how to solve these problems. Have a look at wikipedia here http://en.wikipedia.org/wiki/Vector_%28s... Most of these can be solved by adding the vectors (a bit of geometry could be useful) Good luck !

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Did you like the answer? We found more questions related to the topic: **Wind homework**

Based on the information you gave me, there are two possibilities. The first one is one that was already mentioned, the movement of each thunderstorms along a line of showers and thunderstorms are moving to the northeast and east. But the entire line is moving to the east and southeast. However, if that is the case, it should also be mentioned in the statement or warning. The second case would be thunderstorms are moving northeast to east, but new thunderstorms are forming at the southern end of the line. And the thunderstorm in the northern end of line is weakening. In this case, everything would look like it is moving to southeast, but those southern end of the line are really new storms that are rapidly forming along south end of the boundary or line of storms. This is very common sight to see on radar. Another big clue that this maybe happening would be to check the weather observations to the southeast of the storms. If the winds to the southeast of these thunderstorms are blowing from the southeast, then I would say the second "building new thunderstorm" solution would likely be the reason why (there seems to be a difference from what they are telling you and what you are seeing from your point of view. If the direct phones lines are not open to the NWS and you think the information is critical for them to look into, upi can contact an emergency community official or one of your local media stations that usually broadcast weather alerts and warnings and request that they relay your concerns to the NWS. Also, if you know of someone who is an official weather spotter for the NWS or SKYWARN, you can contact that person and request that he pass on your concerns to the NWS.

The NWS are experts. They can tell which direction a storm is moving. If you think you are right, then don't go by what they say and just go by what you think will happen. I don't know where you live or anything so I have no way of predicting if it will fall apart before it gets to you. You didn't say how strong the storm was or anything. I don't mean to sound rude, but you need to put a little more information for anyone to give there opinion on if it will hit your area. A lot of times, you have a line of storms heading in a direction, but the actual line can be moving in one direction, but the actual storms in the line can be moving in a different direction. Maybe that is what you heard.

Draw a picture and solve the triangles for vector math. due east = 300 + 100 * cos(30) = 386.6 due north = 100 * sin(30) = 50 a^2 + b^2 = c^2 c^2 = (386.6)^2 + 50^2 = 151,961 c = 389 mph

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There is a stipulation that the conservation of momentum will be used in this problem. I will simply outline the steps of the solution of this problem. I will leave the actual algebraic manipulations and calculations up to you. Step 1 -- determine the velocity of the bullet and block as it starts the slide Working formula is Vf^2 - V^2 = 2as where Vf = final velocity = 0 (when the block and bullet stoops) V = velocity of the block & bullet before sliding a = acceleration of the block & bullet s = stopping distance = 1.1 m The above simplifies to 0 - V^2 = 2.2a a = -(V^2)/2.2 --- call this Equation A Step 2 -- using Newton's 2nd Law of Motion, F = ma where F = force acting of the block & bullet = µMg µ = coefficient of friction = 0.40 (given) M = mass of the bullet & block = 10/1000 + m = (0.010 + m) g = acceleration due to gravity = 9.8 m/sec^2 (constant) and substituting values, - 0.40(0.010 + m)(9.8) = ma The negative sign on the force indicates that only the frictional force is acting on the body, which is opposite that of the motion Solving for "a" a = - [0.40(0.010 + m)(9.8)]/m -- call this Equation B NOTE again the negative sign attached to the acceleration. It means that the body is slowing down after its initial slide. Step 3 -- since Equation A = Equation B, then - (V^2)/2.2 = - [0.40(0.010 + m)(9.8)]/m -- Call this Equation C Step 4 -- use the law of conservation of momentum 0.010(292) + m(0) = (0.010 + m)V 2.92 = (0.010 + m)V -- call this Equation D Step 5 -- note that considering Equation C and Equation D, there are 2 equations with two unknowns namely "V" and 'm." I trust that you can solve simultaneous equations. I have outlined the solution and I will leave the tedious mathematical manipulations for you to do. Hope this helps.

Initial energy E = 1/2 m1 V^2 = 1/2 (0.01) kg (292)^ (m/s)^2 = 426.32 J -- (m1 is the mass of the bullet) Using the work-energy theorem this equals the work done by the friction over the stopping distance: E = friction force x distance (*) ; but we know friction = normal force x coefficient of friction, and the normal force is just the weight. Let m2 be the mass of the block, then Friction force = 0.40 (m2 + m1) g = 0.40 x (m2 + 0.01) kg x 9.81 m/s^2 = 0.039 + 3.92 m2 (in Newtons) Hence using (*), 426.32 / 1.1 = 0.0392 + 3.924 m2 -----> m2 = 98.76 kilograms.

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