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Answers toStudent Questions

Physics

What is the gravitational field strength formula?

A gravitational field is a region in space where a mass will experience a gravitational force. The gravitational field strength at a location is defined as the force per unit mass experienced by a small test mass placed at that location. The field strength can be expressed as an equation as follows: $\hspace{3em} g = \dfrac{F}{m}$ Where:\ $\hspace{3em}$ $g$ is the gravitational field strength in units of N kg$^{-1}$\ $\hspace{3em}$ $F$ is the gravitational force experienced by the mass in units of N\ $\hspace{3em}$ $m$ is the mass of the object experiencing the gravitational force in units of kg. Any object with mass will create a gravitational field in the region around it. This field extends an infinite distance away; however, it decreases proportionally with the square of the distance. We can use Newton’s Law of Gravitation to develop the formula for the gravitational field strength at a given distance from an object. The law states that the force of gravity between two masses is directly proportional to the product of the masses and inversely proportional to the square of the distance of the separation of their centres of mass: $\hspace{3em} F_g = G\dfrac{m_1m_2}{r^2}$ Where:\ $\hspace{3em}$ $F_g$ is the force of gravity in units of N \ $\hspace{3em}$ $G$ is the gravitational constant 6.67 x 10$^{-11}$ N m$^2$ kg$^{-2}$\ $\hspace{3em}$ $m_1$ and $m_2$ are the masses of the two objects in units of kg\ $\hspace{3em}$ $r$ is the separation of their centers of mass in m. We can substitute this equation for gravitational force into the first question for gravitational field to derive a new equation $\hspace{3em} g = \dfrac{F}{m}$ $\hspace{3em} g = \dfrac{G\dfrac{m_1m_2}{r^2}}{m}$ $\hspace{3em} g = \dfrac{Gm_1}{r^2}$ This formula gives the gravitational field strength at a distance $r$ from a the centre of a mass $m_1$ . This mass is normally a planet, star or moon.

Biology

How do we determine if genes are autosomal vs sex linked?

To determine whether genes are autosomal or sex-linked, we first need to understand the difference between the two. Autosomal genes are located on the autosomes, which are the 22 pairs of non-sex chromosomes in humans. These chromosomes are present in both biological males and females in equal number, so traits governed by autosomal genes typically show similar inheritance patterns across sexes. In contrast, sex-linked genes are located on the sex chromosomes, primarily the X chromosome. Since biological males have one X and one Y chromosome, and biological females have two X chromosomes, the inheritance of sex-linked traits differs between sexes. Y-linked traits, though rare, are located on the Y chromosome and are passed strictly from father to son, affecting only biological males. To identify whether a gene is autosomal or sex-linked, we look at inheritance patterns by examining how traits are passed down through generations. For human traits these studies often include the use of pedigree charts, which are visual diagrams showing family relationships and the presence or absence of specific traits. Pedigree charts use standardised symbols to represent individuals, their biological sex, and whether they express or carry a trait. By analysing these charts, we can observe patterns of inheritance that help distinguish between autosomal and sex-linked traits. For example, autosomal traits usually affect males and females equally. Recessive autosomal traits can skip generations. In contrast, X-linked traits often appear more frequently in biological males because they have only one X chromosome. A recessive allele on that chromosome will be expressed, whereas biological females need two copies of the recessive allele (one on each X chromosome) to express the trait. If a biological female is a carrier of a recessive X-linked trait, each of her sons has a 50% chance of inheriting the allele and expressing the trait, while each daughter has a 50% chance of being a carrier. Daughters will only express the trait if they inherit the recessive allele from both parents. Classic examples of X-linked recessive conditions include haemophilia (as stated in the IB Biology syllabus subtopic D3.2) and red-green colour blindness, both of which disproportionately affect biological males Further patterns help confirm sex linkage. If a biological male with an X-linked trait has children, all his daughters will inherit the allele (since daughters receive his X chromosome), but none of his sons will (since they inherit his Y chromosome). In contrast, autosomal traits do not show sex-specific inheritance; both sons and daughters have an equal chance of inheriting the allele from either parent. In addition to human studies, genetic crosses in model organisms like fruit flies (*Drosophila*) and mice provide valuable insights into the inheritance of sex-linked and autosomal traits. For example, in fruit flies, X-linked traits like white eye colour appear more frequently in males because they have only one X chromosome, while females need two copies of the recessive allele to express the trait. By performing controlled crosses, scientists can predict inheritance patterns and distinguish between autosomal and sex-linked genes. In summary to determine whether genes are autosomal or sex-linked, we analyse inheritance patterns through tools like pedigree charts. Autosomal genes are located on non-sex chromosomes and are inherited equally by both sexes, while sex-linked genes are found on the sex chromosomes and show different inheritance patterns between males and females. Genetic crosses in organisms like fruit flies and mice help confirm these patterns, providing additional insights into the inheritance of sex-linked and autosomal traits.

Mathematics

What are double angle identities?

Double angle identities appear in both the AA SL and AA HL courses, but do not appear in either the AI SL or AI HL courses. There are double angle identities for sine and cosine, which appear in both the AA SL and AA HL courses, and a double angle identity for tangent, which appears only in the AA HL course. There is one double angle identity for the sine ratio - $\sin 2\theta = 2 \sin \theta \cos \theta$ There are three double angle identities for the cosine ratio - $\cos 2\theta = \cos ^2 \theta - \sin ^2 \theta$ - $\cos 2\theta = 2 \cos ^2 \theta - 1$ - $\cos 2\theta = 1 - 2 \sin ^2 \theta$ There is one double angle identity for the tangent ratio - $\tan 2\theta = \dfrac{2 \tan \theta}{1 - \tan ^2 \theta}$ **The double angle identities give us a way of expressing a trigonometric ratio in another form that may make it easier to solve an equation, or help us to find the exact value of a new angle, particularly in paper 1 where we cannot use a G.D.C. to help us.** For example, suppose we are asked to solve the equation $\sin \theta = \dfrac{1}{4 \cos \theta}$ for $0 \leq \theta \leq 2\pi$. We could rearrange this by multiplying both sides by $2 \cos \theta$, giving us $$ \begin{align*} \sin \theta \times 2 \cos \theta &= \dfrac{1}{4 \cos \theta} \times 2 \cos \theta \\[8pt] 2 \sin \theta \cos \theta &= \dfrac{1}{2} \end{align*} $$ Now on the left hand side we can apply the double angle identity for sine so that we have $$ \begin{align*} \sin 2\theta &= \dfrac{1}{2} \end{align*} $$ From here we can proceed as though solving a “normal” trigonometric equation, finding that the two solutions are $\theta = \dfrac{\pi}{12}$ and $\theta = \dfrac{5\pi}{12}$. As a second example, suppose we are asked to find the exact value of $\tan \left(\dfrac{\pi}{12}\right)$. From ratio triangles, we know that $\tan \left(\dfrac{\pi}{6}\right)=\dfrac{1}{\sqrt 3}$ and $\dfrac{\pi}{6}$ is double the angle $\dfrac{\pi}{12}$ so we could use the double angle identity for tangent to help us. We can set up our equation as follows: $$ \begin{align*} \tan 2\theta &= \frac{2 \tan \theta}{1 - \tan ^2 \theta} \\[12pt] \tan \left(\frac{\pi}{6}\right) &= \frac{2 \tan \frac{\pi}{12}}{1 - \tan ^2 \frac{\pi}{12}} \\[12pt] \frac{1}{\sqrt 3} &= \frac{2 \tan \frac{\pi}{12}}{1 - \tan ^2 \frac{\pi}{12}} \end{align*} $$ Now we could rearrange this to make a quadratic equation. $$ \begin{align*} \tan ^2 \left(\frac{\pi}{12}\right) + 2\sqrt{3} \tan \left(\frac{\pi}{12}\right) - 1 = 0 \end{align*} $$ We could then complete the square, or use the quadratic formula to find that $\tan\left(\dfrac{\pi}{12}\right) = \pm 2 - \sqrt {3}$. Then we can reason that $\dfrac{\pi}{12}$ is in quadrant 1 where the tangent function is positive, hence $\tan\left(\dfrac{\pi}{12}\right)$ must equal $2 - \sqrt{3}$. If faced with a trigonometric equation or expression in an exam that doesn’t appear to be solvable or easy to simplify in its current form,, it is useful to investigate whether a substitution using a double angle identity makes the equation more easily solvable, or the expression more easy to simplify.

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