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2 years ago

Calculate the couple moment acting on the steering wheel, and express it as a Cartesian vector.

Mathematics

1 answer:

IRISSAK [1]2 years ago

7 0

The equation for resolving couples will come in place

$Fcos\theta=0\\\ Fsin\theta =0$

Using the unit vectors by i and j along the x-axis and the y-axis you'll express it as a Cartesian vector.

<h3>Couple moment and Cartesian vector</h3>

Couple moment:This looks at the product of the forces acting on a body( in this context a steering wheel ) by the perpendicular distance that exist amidst the forces.

Cartesian vector:this is the representation of unit vectors by i and j along the x-axis and the y-axis, respectively.

Therefore, in calculating the <em>couple</em> moment t acting on the steering wheel, the equation for resolving couples will come in place

$Fcos\theta=0\\\ Fsin\theta =0$

Using the unit vectors by i and j along the x-axis and the y-axis you'll express it as a Cartesian vector.

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Evaluate the expression shown below and write your answer as a fraction in simplest form 1/9+1/12

netineya [11]

The answer would be 7/36.

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3 years ago

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3 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago

Step by step explanation please

galina1969 [7]

Two equal triangles

2(1/2×b×h)

=2(1/2×8×10)

$= 80 {cm}^{2}$

Three rectangles

=3(l×h)

=3(18×8)

=3(144)

$432 {cm}^{2}$

Total Surface area

$80 {cm}^{2} + 432 {cm}^{2} \\ 512 {cm}^{2}$

8 0

3 years ago

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Crazy boy [7]

Answer:

Non-linear because none of them repeat.

Step-by-step explanation:

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3 years ago

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