When working with vectors, you will often see right triangles. What are the consistent properties of these triangles?

Physics

1 answer:

german3 years ago

4 0

Answer:

a) and c)

Explanation:

Any vector can be expressed as a vector sum of its x- and -y components, as follows:

v = vₓ* x + vy* y

where vₓ = x- component, x= unit vector in the x direction, vy = y-component, y = unit vector in the y direction.

If we add the components graphically, using the head-to-tail method, it will be defined a right triangle, being the vector sum of both components (the vector itself) the hypotenuse of the triangle.
As both components are perpendicular, they will be always lined up with the x- and y- axes.

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#1 Not sure where to start. This is for AP Physics!

yaroslaw [1]

First,

$\rho=\dfrac mV$

where $\rho$ is density, $m$ is mass, and $V$ is volume. We can compute the volume of the roll:

$2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V$

$\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3$

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness $x$ . Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

$V=Ax$

where $A$ is the given area, so

$4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x$

$\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}$

If we're taking significant digits into account, the volume we found would have been $V=470\,\mathrm m^3$ , in turn making the thickness $x=250\,\mathrm{mm}$ .