الفصل الثالث(فيزياء): المتجهات

ALI ALZAHRANI · 30-10-2009, 08:53 PM

المشاركة الأصلية كتبت بواسطة نوف18

الله يجزاك الجنه ويوفقك يآآآآآآآآآآآآآآآآآآآآآآآآآآرب

عندي اسئله نقرتين لعرض الصورة في صفحة مستقلة

×××

1) the x component of vector A is -25.0 m and the y component is + 40.0 m
(a) what is the magnitude vector A ?
The magnitude of the vector A is given by:

A = (Ax^2 + Ay^2)^0.5=([-25]^2+40^2)=47.2 m

(b) what is the angle between the direction of vector A and the positive direction of x ?

The angle the vector A makes with x-axis is given by:

Ax = AcosQ

therefore the angle is

Q=cos-1(Ax/A)=cos1-(-25/47.2)=121.98
×××

3) what are (a) the x component and (b) the y component of a vector a in the xy plane if its direction is 250 counterclock wise from the positive direction of the x axis and magnitude is 7.3 m?

The x-component of the vector A is given by:

Ax = AcosQ=7.3 cos250=-2.5

The y-component of the vector A is given by:

Ax = AsinQ=7.3 sin250=-6.85

×××

6)a displacement vetor r in the xy plane is 15 m long and directed at angle ∂=30 in fig 3-29.determine (a) the x component and (b) the y component of the vector.

The x-component of the vector r is given by:

rx = r cosQ=15 cos30=13.0

The y-component of the vector r is given by:

rx = r sinQ=15 sin30=7.5
×××

13 ) two vectors are given by
a= (4.0m)i - (3.0m)j = (1.0m)k
b= (-1.0m)i - (1.0m)j = (4.0m)k
in unit-vector notation , find (a) a+b,(b) a - b and (c) a third vector c such that
a - b + c=0

a+b = 5i - 2j - 3k

a+b = 3i - 4j +5k

a-b+c=0 ======> c= b-a = -5i + 2j +3k

×××

17) the two vector a and b in fig 3-30 have equal magnitudes of 10.0 m and the angles are ∂1=30 and ∂2=105.
find the (a) x and (b) y components of their vector sum r , (c) the magnitude of r, and (d) the angle r make with the positive direction of the x axis

ax = a cosQ1=10 cos30=8.66
ay = a sinQ1=10 sin30=5.0

bx = b cosQ2=10 cos135=-7.07
by = b sinQ2 =10 sin135=7.07

rx = ax+bx=8.66-7.07=1.59
ry = ay+by=5.0+7.07=12.07

The magnitude of the vector r is
r = (rx^2 + ry^2)^0.5=(1.59^2+12.07^2)=12.2 m

The angle the vector r makes with x-axis is given by:

rx = r cosQ

therefore the angle is

Q=cos-1(rx/r)=cos1-(1.59/12.2)=82.5
×××

Please make another iteration to realise the calculation

نوف18 · 30-10-2009, 10:30 PM

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