Using this equation,a1v1=a2v2,explain how jet engines provide a forward thrust for an airplane.?
342165
This may actually have to do with Newtons laws of physics. This one would seem to be "Every action has an equal and opposite reaction"
312297
For this you should know that sound requires a medium to travel and is facilitated by wavwfront formation. These wavefronts (or disturbances) are in form of varying pressure inside the medium (in our case air). The disturbances (or waves) in form of high and low pressure zone travels at what is called the speed of sound in the air originating from source (the Jet engine in this case).\n\nNow when some jetplane approaches the speed of sound it is acually trying to travel with the high and low pressure zones created by himself. Along the length of the aircaraft itself there is a lot variation in air pressure. This disturbs whole of the aerodynamics of the plane and it becomes very difficult to overcome the resulting drag.\n\nThis is what is called a sound barrier
305602
The reaction is\n\nBa+2 + 2F- <==> BaF2\n\nLet x = [Ba+2] and [F-] = .1 M\n\nQ = [Ba+2][F-]^2 = .01x\n\nA precipiate forms when Q > Ksp.\n\nThis happens when .01x > 2*10^-6\nx = [Ba+2] = 2*10^-4\n\nYeah, sorry, I got confused for a second.  There's a few listings here if you're interested: http://www.towson.edu/~ladon/solprod.html
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Mach is the unit of calculation for speed, whereby its how fast a sound wave can travel in 1 second.\n\nHence, Mach 1 means the 1 times the speed of sound, which is the basic unit of measurement. Mach 2 henceforth means twice the speed of sound and so on.\n\nModern military jets usually have a published speed of Mach 1 to Mach 2. However, that's the given speed when the aircraft is in certain configuration, for instance a clean configuration where there are no external stores like missiles or drop tanks (fuel). A fighter jet will rarely break the Mach 1.8 limit in a combat configuration (with external ordnance like missiles, drop tanks, or bombs).\n\nThe fastest speed will therefore depend largely on the aerodynamic shape of the jet in question, the role of the jet, the type and efficiency of its engines, and whatever is carried externally on the wings.\n\nAs a general rule, pilots rarely go supersonic as it depletes the fuel reserves quickly.
890224
Velocities are vectors, so you´ll have to proceed according to vector rules for addition/subtraction.  Draw a horizontal line, with an arrowhead pointing to the left. This represents the (unknown) velocity due West. Now draw another line, vertical, this time.  Put an arrowhead in the lower end of this line.  This stands for the 50 kph wind speed to the South.  These two lines should make a right (90°) angle.  The lines haven' t to be drawn to scale.\n\nFinally, draw an oblique line connecting both remaining ends.  This is the aircraft speed; to indicate direction, put an arrowhead  at its leftmost end.\n\nNow, you have a triangle of which two sides are known.  This, and the fact it is a right triangle, suffices to solve the entire triangle. The angle you seek is that formed between the horizontal and oblique lines.  Let θ denote this angle.  From elementary trigonomety, sin θ = opposite side/hypotenuse = 50/205 = 0.2439... All you have to do now is to find the angle whose sine is 0.2439...  This is 14.1169...°.  From the figure you have drawn, this angle lies N of W.\n\nThere are several ways to figure out the speed to ground.  You may use Pythagoras' Theorem; or, having found the opposite angle, you may compute it as 205 × cos 14.117° = 198.809...\n\nJust follow the same procedure to solve the other problem.  You'll have to decide which direction is "river across" and which is "downstream"; I´ll suggest "right" and "down", respectively.  Just plug in the proper figures.
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Newtons second law - blance of momentum.\n\nWhen the force acting on the mass to accelerate it is the maximum that the engine, atmosphere, and gravitational attraction to the earth can produce, and those forces are balanced by the the forces acting on the mass to slow it down, then the mass is at its "max speed".\n\nDensity of the air is affected by altitude, temperature, humidity, and weather.  For horizontal, noncompressible flight, that density is a major component of the "dynamic pressure" and therefore the drag.\n\nA very simple model of drag is D=0.5*rho*U^2 *Cd*A where rho is the density of the working fluid (atmosphere), U is the freestream speed of the air, Cd is the coefficient of drag of the airplane, and A is the frontal cross-sectional Area.\n\nIf you set max thrust equals this drag, then you can math out what U is, and call it the max speed of the plane.  If the plane has a headwind or tailwind then its absolute speed will vary.\n\nIf the plane approaches mach then stranger and more energy-demanding phenomena kick in.  If the plan is going beyond the speed of sound there is a third, also complex description for the drag.\n\nIf you are using a prop driven plane then you cant just assume that the thrust equals the planes horsepower, you have to consider the efficiency of the prop and there are equations for that.\n\nDetermining the Cd for a plane is complex because the wing, body, and tail all interact to change the overall Cd.
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The efficiency of an ideal engine is (1- T2/T1) or (T1- T2)/T1 = (600 - 400) / 600 =0.333.\nBut in the problem it is given that it is 0.5 which is not possible.
840090
P1V1 = nRT = P2V2\nP1/P2 = V2/V1
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I think it is 225mph, since with a tailwind(2w) the plane goes 300mph, and with a head wind(-2w) the plane goes 150mph.\n300mph - 150mph = 150mph difference\n150mph dif. / 2 =75mph as the midpoint between 300 and 150\n300 - 75 = 225 and 150 + 75 = 225
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CW= PW / (.5)(A)(Gr)(vcubed)\n\nCW=drag coefficient\nPW= aerodynamic power (kilowats)\n.5= constant\nA= total front area of the vehicle in meters squared.\nGr= gradient of the terrain.\nv3= velocity cubed.\n\nGr can be found with the equation:\nGR=(mass)(gravity)(velocity)(sin of the angle of the gradient)\n\nyour total answer should be  no more than 1
921196
If you run x1 meters against wind, the work you produce is A1=F*x1+f*x1; F is the force of your muscles to overcome the resistance of air with no wind, f is the force of the wind in your face. Let x1 be covered in time t1, then P1=A1/t1 is your power against the wind. \nIf you run x2 meters with the wind, the work you produce is A2=F*x2-f*x2; F & f being the same.\nLet x2 be covered in time t2, then P2=A2/t2 is your power with the wind. They say that dP=P1-P2=50W or dP=(F+f)*(x1/t1)-(F-f)*(x2/t2). As x1/t1=x2/t2=v, then dP=2*f*v. Thus f=dP/(2*v)=50/7=7.14N  is the force of the wind, m=80kg is excessive datum!
871212
i2=i1R1/R2\ni1=R2i2/R1\n\nR1/R2=i2/i1\n\nR1/Rx = R2/R3\n(R1/R2)*R3=Rx\nRx=(i2/i1)*R3
808209
Using one of the equations of motion\n\n(v*v) - (u*u) = 2*a*s\n\nwhere \nv is final velocity\nu is initial velocity\na is accleration\ns is displacement\n\n\nOR \nuse \n\nv * (dv / ds) = a
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let V1= 4 mph, V2= 5 mph\n\nV1*t + V2*t = 3\nt= 20 min\n\n19. a 20 min\n\n\n14.16/3 = 4.72\n\n4.72*14 = 66.08\n\n20 c 66.08
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Since the plane is horizontal normal force cancel weight so no force in vertical direction. No friction so no extra force in the horizontal direction.\nF = ma System: wagon\nFn + Fg + F1 + F2 = ma   \n(these are all vectors, except the m, but I can't type the arrow)\nF1 + F2 = ma (Fn + Fg = 0) (again, vectors)\nChoose the 45N as positive direction\n45 - 30 = 8a\n15 = 8a\na = 1.875 m/s^2 in the 45N force direction
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since P1.V1 / T1 = P2.V2 / T2 ... u don't actually care about the altitude. so V2 = 8.88 cm^3 (btw those degrees are °C, °F ?? cuz u gotta take them to Kevin before using them)\n\n1*10^5 Pa * 2cm^3 / 270 °K = 2*10^4 Pa * V2 / 240 °K\nV2 = 8.88 cm^3
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a. Applying conservation of momentum\nm1 x V_i + m2 x 0 = (m1+m2) x V_f\nV_f = (m1 x V_i)/(m1+m2)\n\nb. If m1 >> m2 then m1 + m2 -> m1\nTherefore V_f = V_i\n\nc. If m1 << m2 then m1+m2 --> m2\nTherfore V_f = (m1/m2) x V_i\n\nd.m1 x V_i - m2 x V_i = (m1+m2) x V_f\n=> V_f = ((m1-m2)x(V_i))/(m1+m2)\n\ne. m2 = m1, m2 =(1/3)(m1)