WER Câu 11. Ôtô bắt đầu chuyển động từ trạng thái đứng yên và đạt đến tốc độ c 70 ft/s) sau khi đi được quãng đường d – 400 ft)
1 answer:
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First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:
1) Take first force, say
, and assume that it is pointing towards the x-direction.
Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
= 
, where 
= Unit vector in the x-direction.
2) Take the second force, say
, and assume that it is making an angle 
with the first force .
Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:
Now from simple vector addition, we know that,
--- (A)
Where
= Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.
Let us find
first!
= 
=> = 
Now the magnitude of is,
|| = 
=> || = 
Since
, therefore,
=> || = 
Since || = ||, and the magnitude of the resultant force is 20N, therefore,
|| = | |
20 =
Take square on both sides,
400 =
Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°
-israr
1) Take first force, say
Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
2) Take the second force, say
Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:
Now from simple vector addition, we know that,
Where
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.
Let us find
=>
Now the magnitude of
|
=> |
Since
=> |
Since |
|
20 =
Take square on both sides,
400 =
Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°
-israr
Answer:a) 629,5851 km/h in magnitude b)629,5851 km/h at 84,2 degrees from east pointing south direction or in vector form 626,6396 km/h south + 63,6396km/h east. c) 16,5 km NE of the desired position
Explanation:
Since the plane is flying south at 690 km/h and the wind is blowing at assumed constant speed of 90 km/h from SW, we get a triangle relation where
see fig 1
Then we can decompose those 90 km/h into vectors, one north and one east, both of the same magnitude, since the angle is 45 degrees with respect to the east, that is direction norhteast or NE, then
90 km/h NE= 63,6396 km/h north + 63,6396 km/h east,
this because we have an isosceles triangle, then the cathetus length is
hypotenuse/
using Pythagoras, here the hypotenuse is 90, then the cathetus are of length
90/ km/h= 63,6396 km/h.
Now the total speed of the plane is
690km/h south + 63,6396 km/h north +63,6396 km/h east,
this is 626,3604 km/h south + 63,6396 km/h east, here north is as if we had -south.
then using again Pythagoras we get the magnitude of the total speed it is
,
the direction is calculated with respect to the south using trigonometry, we know the
sin x= cathetus opposed / hypotenuse,
then
x= =5,801 degrees from South as reference (0 degrees) in East direction or as usual 84,2 degrees from east pointing south or in vector form
626,6396 km/h south + 63,6396km/h east.
Finally since the detour is caused by the west speed component plus the slow down caused by the north component of the wind speed, we get
Xdetour{east}= 63,6396 km/h* (11 min* h)/(60 min)=11,6672 km=Xdetour{north} ,
since 11 min=11/60 hours=0.1833 hours.
Then the total detour from the expected position, the one it should have without the influence of the wind, we get
Xdetour=[/tex]\sqrt{2* 11,6672x^{2} }[/tex] = 16,5km at 45 degrees from east pointing north
The situation is sketched as follows see fig 2
