In a right triangle $$ABC$$, given that $$cos\angle BAC=3/4$$, and $$BC=7cm$$, what is the length of the hypotenuse, $$AB$$?
$$AB=4\sqrt7cm$$. Since we know that $$\triangle ABC$$ is a right angled triangle, and $$AB$$ is the hypotenuse, by drawing a diagram we know that $$cos\angle BAC= AC\AB$$. Using pythagoras theorem, we can deduce that $$sin\angle BAC=BC/AB=\sqrt (AB^2-AC^2)/AB=\sqrt7/4$$. By substituting the value for $$BC$$, we get that $$7/AB=\sqrt7/4$$. Cross multiplying gets us $$AB=\sqrt7/4cm$$.
$$4NO_2+$$_$$H_2\rightarrow$$_$$NH_3+$$_$$H_2O$$ What is the sum of all the numbers in the blanks?
26. This is a simple equation balancing question. One method to do this is to identify that on the RHS, only $$H_2O$$ has oxygen, so there must be 8, based on the number of oxygen in $$4NO_2$$. Then notice that on the RHS, only $$NH_3$$ has nitrogen, so there must be 4. Finally, the number of hydrogen on the LHS is equal to the number of hydrogen on the RHS, which comes 14$$H_2$$ atoms. All the blanks then add up to 26, the final answer.
Two objects, A and B, are in motion in the same direction and have masses $$m$$ and $$3m$$ respectively. A has velocity $$2v$$ and B has velocity $$v$$. After colliding and sticking together, what is their new kinetic energy?
$$25/8mv^2$$. This question is about momentum and energy. After A and B become one object, they have a total mass $$4m$$. Since momentum is always conserved, the combined object must have momentum $$5mv$$, which is the sum of A and B's momentums. We can then deduce that the new object has velocity $$5/4v$$. By using the equation $(KE=1/2mv^2$) And substituting the values of $$m$$ and $$v$$ for the combined object, the kinetic energy of the new object is $$25/8mv^2$$.