Physics 2011

A cross product is always perpendicular to both of its factors, so $\vec{A}\cdot\vec{C}=0$ holds for any $\theta$; and $-\vec{C}=\vec{B}\times\vec{A}$ and $|\vec{C}|=|\vec{A}||\vec{B}|\sin\theta$ are the definition of the product itself, so they hold too.

The dot product is the different case: $\vec{A}\cdot\vec{B}=0$ requires the two vectors to be perpendicular. The question fixes $\theta\neq90^\circ$, so that is the statement which fails.

$v_0=20\,\mathrm{m/s}$, $v=0$, $t=2.0\,\mathrm{s}$

The car decelerates uniformly to rest, so the skid distance is the average velocity times the time, $d=\bar v\,t=\dfrac{v_0+v}{2}t$. $$d=\frac{20+0}{2}\times2.0=20\,\mathrm{m}$$

Displacement is the area under a velocity-time graph.

Adding the triangle from the origin to B, the trapezoid from B to C, the rectangle from C to D and the triangle from D to E: $$4+8+36+36=84\,\mathrm{m}$$ Between D and E the slope is negative rather than positive, and between B and C it is $8\,\mathrm{m/s^2}$ rather than $4\,\mathrm{m/s^2}$, so the total displacement of $84\,\mathrm{m}$ is the only true statement.

At the bottom of the loop the rope must supply the centripetal force and support the object's weight, while at the top gravity itself contributes to the centripetal force: $$T_{bottom}=\frac{mv^2}{r}+mg,\qquad T_{top}=\frac{mv^2}{r}-mg$$

The object is also fastest at the bottom, by conservation of energy, so both effects favour the bottom — the tension is greatest there.

$\vec{F}_1=3\hat{\imath}\,\mathrm{N}$, $\vec{F}_2=4\hat{\jmath}\,\mathrm{N}$

A force that balances the two is equal in magnitude and opposite in direction to their resultant $\vec{R}=\vec{F}_1+\vec{F}_2$. $$R=\sqrt{3^2+4^2}=5\,\mathrm{N},\quad\text{at }\tan^{-1}(4/3)=53^\circ\text{ counterclockwise from }\vec{F}_1$$ The balancing force $\vec{F}_{eq}$ therefore also has magnitude $5\,\mathrm{N}$, pointing opposite $\vec{R}$ — which is $53^\circ+90^\circ=143^\circ$ counterclockwise from $\vec{F}_2$.