#1

Some questions i need help with. I don't care about the answer (i already know them as well), i wanna know how to do them.

A car goes around a curved stretch of flat roadway of radius R = 110.0 m. The magnitudes of the horizontal and vertical components of force the car exerts on a securely seated passenger are, respectively, X = 210.0 N and Y = 600.0 N.

The first two parts of the question then ask for i) the speed and ii) minimum coefficient of static friction between the tyres and the road needed to negotiate this turn without sliding out

i) is simply F=(mv^2)/r where F = 210N and m = Y/9.81=61.16kg

which gives v=19.43m/s = 70km/hr

ii) is just X/Y using the old max force = mew (cooefficient of static friction) x normal force. So it's 0.35

Now the part i can't seem to get:

iii) This stretch of highway is a notorious hazard during the winter months when it can be quite slippery. Accordingly the LTSA decides to bank it at an angle φ = 23.0 ° to the horizontal. At what speed could the car now negotiate this curve without needing to rely on any frictional force to prevent it slipping upwards or downwards on the banked surface?

Currently i've got a conceptual problem with visualizing the forces acting on/by the car. The answer is 77km/hr but like i said i want to know how to get it.

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Next question:

To identify people suspected of being infected with SARS, airports around the world have set up an infrared cameras that measures the infrared radiation that people give off. What is the percentage increase in the rate of heat radiated from a person with a surface skin temperature of 35.0 °C compared with the same person with a skin temperature of 33 °C?

Now, the relevant equations (i think) are:

dQ/dt (heat transfer) = -k.A.(dT/dx) where k =thermal conductivity, A = area

and dT/dx = rate of change of temperature with respect with thickness, which is equal to (T[hot] - T[cold])/ l where T[hot]>T[cold] and l is the thickness.

Now, what i figure is that seeing as how we can take k, A, l, and T[cold] to be the same, it should just be a matter of finding the temperature difference as a percentage, but that doesn't seem to be the case?

The answer is 2.64%

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Final one (i won't ask you guys the other one ):

A fully loaded Cessna-182 airplane of mass 1290 kg has an engine failure when flying with an airspeed of 129 km/h at an altitude of 2720 m on a calm day. It then glides at a constant glide angle (which is the direction of flight below the horizontal) towards a safe landing at this constant speed of 129 km/h experiencing a drag force of 1280 N that opposes the direction in which the plane is moving.

Then there are a few small parts unrelated to the next part...

Then it says:

Suppose the pilot instead had managed to get the airplane engine started such that he was able to apply full throttle and the airplane climbed along a straight line angled above the horizontal so that it gained altitude at a steady rate of 5.63 m/s. Assuming he was again flying with an airspeed of 129 km/h determine:

i)The flight angle above the horizontal the plane is flying

ii)The thrust force of the engine (that acts in the direction the plane is moving)

i) is fine, it's simply the inverse sine of (new v in y direction / v) and is 9.04 degrees.

Part ii) i absolutely don't seem to get. I tried drawing the force diagram and then expressing each force in vector components then equating vertical and horizontal. Didn't work

The answers 3270N.

C'mon fellow physicists, help a brother out

A car goes around a curved stretch of flat roadway of radius R = 110.0 m. The magnitudes of the horizontal and vertical components of force the car exerts on a securely seated passenger are, respectively, X = 210.0 N and Y = 600.0 N.

The first two parts of the question then ask for i) the speed and ii) minimum coefficient of static friction between the tyres and the road needed to negotiate this turn without sliding out

i) is simply F=(mv^2)/r where F = 210N and m = Y/9.81=61.16kg

which gives v=19.43m/s = 70km/hr

ii) is just X/Y using the old max force = mew (cooefficient of static friction) x normal force. So it's 0.35

Now the part i can't seem to get:

iii) This stretch of highway is a notorious hazard during the winter months when it can be quite slippery. Accordingly the LTSA decides to bank it at an angle φ = 23.0 ° to the horizontal. At what speed could the car now negotiate this curve without needing to rely on any frictional force to prevent it slipping upwards or downwards on the banked surface?

Currently i've got a conceptual problem with visualizing the forces acting on/by the car. The answer is 77km/hr but like i said i want to know how to get it.

_____________________________________

Next question:

To identify people suspected of being infected with SARS, airports around the world have set up an infrared cameras that measures the infrared radiation that people give off. What is the percentage increase in the rate of heat radiated from a person with a surface skin temperature of 35.0 °C compared with the same person with a skin temperature of 33 °C?

Now, the relevant equations (i think) are:

dQ/dt (heat transfer) = -k.A.(dT/dx) where k =thermal conductivity, A = area

and dT/dx = rate of change of temperature with respect with thickness, which is equal to (T[hot] - T[cold])/ l where T[hot]>T[cold] and l is the thickness.

Now, what i figure is that seeing as how we can take k, A, l, and T[cold] to be the same, it should just be a matter of finding the temperature difference as a percentage, but that doesn't seem to be the case?

The answer is 2.64%

___________________________________________________

Final one (i won't ask you guys the other one ):

A fully loaded Cessna-182 airplane of mass 1290 kg has an engine failure when flying with an airspeed of 129 km/h at an altitude of 2720 m on a calm day. It then glides at a constant glide angle (which is the direction of flight below the horizontal) towards a safe landing at this constant speed of 129 km/h experiencing a drag force of 1280 N that opposes the direction in which the plane is moving.

Then there are a few small parts unrelated to the next part...

Then it says:

Suppose the pilot instead had managed to get the airplane engine started such that he was able to apply full throttle and the airplane climbed along a straight line angled above the horizontal so that it gained altitude at a steady rate of 5.63 m/s. Assuming he was again flying with an airspeed of 129 km/h determine:

i)The flight angle above the horizontal the plane is flying

ii)The thrust force of the engine (that acts in the direction the plane is moving)

i) is fine, it's simply the inverse sine of (new v in y direction / v) and is 9.04 degrees.

Part ii) i absolutely don't seem to get. I tried drawing the force diagram and then expressing each force in vector components then equating vertical and horizontal. Didn't work

The answers 3270N.

C'mon fellow physicists, help a brother out

#2

I'd like to help, but this far more advanced than I am.

But I shall bump this so that darkstar, Lord Of Physics and Advanced Mathematics (if he's online) shall see it.

But I shall bump this so that darkstar, Lord Of Physics and Advanced Mathematics (if he's online) shall see it.

#3

I'd like to help, but this far more advanced than I am.

But I shall bump this so that darkstar, Lord Of Physics and Advanced Mathematics (if he's online) shall see it.

Thanks dude. There's currently no real rush but i still wanna be able to do them before sort of a couple days so i don't fall behind.

Both darkstar and Yakult helped me about a month ago with a couple problems as well which was really good.

#4

alright i'll help you with the first one, i'm pretty sure its the parallel component of the Fg that is replacing the friction force so the mass stays the same you can calc the Fg and then use trig and the given angle to find the parallel component which should equal the parallel component of the centripetal force, i could be wrong this is all head work, i didn't write anything but it should get you started at least.

#5

damn, what level physics is this, (year of school - college ?)

#6

i'm in 11th grade and taking AP Physics AB but we only lightly covered heat transfer and things of that sort so i could only help with the first, the last on the other hand i can only partially create in my head, at 130 AM : )

#7

oh wait i think i get the third one now,

ok so

the thrust force has to oppose the drag force horizontally and the gravitational force vertically, and theres no acceleration any direction so its just the resultant of a force with those x an y components , i think , double checking

ok so

the thrust force has to oppose the drag force horizontally and the gravitational force vertically, and theres no acceleration any direction so its just the resultant of a force with those x an y components , i think , double checking

#8

that answer of 3270 N can't be right because in order for it to climb it would have to at least overcome gravity which a mass of 1290kg would = a vertical force of 12642 N which would mean the thrust force would have to be greater than 12642 N ,.. dunno what to say ... sorry ,,.... but what i said before about the horizontal component = drag and vertical=gravity , that seems right to me.

#9

i just failed a college physics test today. thanks for reminding me :-x I shall offer no help, because I feel worthless in the subject right now. Luckily teacher drops lowest test score.