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[coffeechocolat]

please help me nha, me phai due on monday.It's physic's 11
1, two boathouses are located on ariver, 1.0km apart on the same shore. Two men make round trips from one boathouse to the other and back. One man paddles a canoe at a velocity of 4 km/h relative to the water, and the other walks along the shore at a constant velocity of 4 km/h. The current in the river is 2.0 km/h in the starting direction of the canoeist.
a, How much sooner than the walker does the canoeist reach the second boathouse?
b, how long does it take each to make the round trip?

2, An ocean liner is streaming at 18 km/h dye south. A passenger strolling on the deck walks towards the rear of the ship at 3 m/s. After walking 12s, he turns right and walks at the same speed towards the rail, 15m from his turning point.
a, what is his velocity, relative to the water, while walking towards the rear? While walking toward the rail?
b, Draw a scale vector diagram, or make a sketch and use the mathematical approach to determine his resultant displacement relative to the water at the end of his walk?

3, Suppose you are being carried along by the wind as you ride in a hot air ballon. The wind is 40 km/h(south). You hold up a smal flag. In which direction wil the flag point

thank you very much

[CAptain DaLaZBoi]

1) a- cause of the water.... u can say that the vel for paddleman hhehehe is 6km/hr .. that meant in 1/6 hr = 10 mins... he reaches the boathouse. while the other guy reaches in 1/4 = 20 mins.
1) b-however when travel back..... he paddleman bi. current ca?n lai nen co`n co' vel = 2km/hr. Therefore.. they will both be reaches back to the first boathouse after 40 mins.

One is enough.... tui ddi lam` hw cua tui

[Chris-T]

#2) 18km/s = 18*1000/3600 = 5 m/s South (or vector -5y)
Walking toward rear (draw +3y) at 3m/s North ==> 2 m/s South
as seen by tht still water (displacement vector drawn as -2y).
While walking (after turning right) to side (East) ==> +3x (vector)
==> 3x - 5y as seen by water. (hope you can visualize the vectors
on x-y coordinates).

[yeudoi]

Thanks For Chris-T input about my mistake here is the rework:



1. T(boat) = 1/(4+2)
->T(boat) = 1/6 hr

T(man) = 1/4 hr

a) T(different) = 1/4 - 1/6
= 1/12 hr
= 5 min

Hence 5 min sooner.

b(i) Boat
T(boat) = 1/(4-2) + 1/6 = 2/3 hr = 40 min
(ii) Man
T(man) = 1/4 + 1/4 = 1/2 hr = 30 min

2)
a(i) Relative Vel. while walking toward the rear.
18*1000/3600 - 3 = 2 m/s due south

a(ii) Relative Vel. while walking toward the rail.
18 Km/h (5m/s) in the y direction and 3 m/s in the x direction <-- Answer


METHOD 2: this is actually the anwer for 2(b), You can write the answer in this form below:

Solve for Velocity Magnitude and direction:

For direction: In this case
tan(theta) = 3/2
-> theta = 56 degree 20'

Pythagorus in this case since velocity are right triangular shape.

V^2 = 2^2 + 3^2
--> V = 3.61m/s

Answer: 3.61 m/s, 56 degree 20' ... south east or west I can't remember the blood... direction. :15:

Check my answer for error, the direction and unit conversion.

2b) See above for mathematic approach: Or you can use a ruler and draw the velocity components to scale by:

draw Velocity vector in the y direction pointing down 5 - 3 = 2 units
draw Velocity vecot in the x direction pointing to the right at 3 units

After that use a protractor and a ruler to find out the Magnitude of the resultant velocity and the direction.

GL.

3(c) In theory if the velocity of the Balloon is equal to the wind then the flag will not fly in any direction.

If the wind velocity is greater than the balloon velocity then the flag will fly in the wind direction.

--
Don't blame me for any mistake :D

[coffeechocolat]

cam on cac ban nhieu lam, nhung ma minh con mot it homwork nua , giup minh nha,thanks
1, parachutists need to maximize their cross-sectional area, whereas the automobile industry wishes to reach a minimum effective area. Explain
2, transport trucks have air diverters on the roof of the truck cab. How do the divers reduce wind drag?
3, What effective maximum drag force do the parachutes have?
4, why the parachutist appears to move upward relative to the camera position?

[Chris-T]

yeudoi: the explanations for problem-2 is OK but you mixed the
units together (km/h & m/s) in getting the numerical solution.
Need to convert them to both m/s or km/h before doing the
vector/calculations.

Also, #1b asked for time for total Round-trip (not just the going
back portion) ==> Need to sum up forward & backward trips
T(walk)=1/4 + 1/4 = 1/2hr = 30min.
T(boat)=1/6 + 1/2 = 10min + 30 min. = 40min.

[yeudoi]

Opps... I didn't see the unit. Hehehehe

Thanks Chris_T I will not stay up late anymore... :15:


Anyway. I have updated the work. Check for it above.

Please recheck the answer and my mistake.

I'm the kind of person who never like to read to much. :15:

[yeudoi]

1. Parachute to have maximum cross section area to have maximum aerodynamics drag force so that parachuter falls down slower.

Automobile to have minimum coss sectional area to have minimum aerodynamic drag so that automobile goes faster.

Draw a vector diagram would be good.

2. Dive have bends to reduce drag. As we knows that drag force acting normal to the surface, by having bends the we minimizing the drag. But in exchange we increase the down force.

3. F = 1/2*rho*A*V^2 (Cd)

Cd = drag coefficent for shape. For maximum I guess Cd = 1 <---- You should check on this.

V terminal velocity <--- I guess this too. have you find out this velocity?

A = Area of a circle
rho = Air density


4. Is the camera attach with him?


Sorry for my careless on the first post

[coffeechocolat]

thanks with your help