# Tutor profile: Tara W.

## Questions

### Subject: Physics (Electricity and Magnetism)

A solenoid with n number of turns and a constant current I in the $$ \phi $$ direction has it's axis the z axis. A non-conducting circular ring with a distributed charge q and a radius larger than the solenoid is placed around the solenoid, where the ring's center also lines up with the z axis. If the current in the ring is reduced, what happens tot he ring?

If the current in the solenoid is reduced it will create a changing magnetic or B field. This will put a torque on the ring, and since it has a charge, yet is non conducting, and therefore the charges in the ring are unable to move throughout the ring, the torque will cause the ring to start spinning.

### Subject: Biology

Having orange skin is a dominant trait and having purple skin is recessive. If a man heterozygous for this gene marries a woman with purple skin, what is the probability that both of their children have purple skin?

Lets label this gene as corresponding to the letter O, where the dominant allele will be represented by O and the recessive will be represented by o. Since the father is heterozygous for the gene this means that he has, and expresses the dominant allele O, but is also a carrier of the recessive allele o. Therefore his genotype is Oo. His wife who has purple skin can only have two recessive alleles or a genotype of oo because if she had a dominant allele she would express the phenotype of orange skin. To complete this question I would recommend creating punnet squares, where we would find that the four possible genotype outcomes for this couples offspring are, Oo, Oo, oo, and oo, with a 1/4 chance of each. Or adding the probability of the same genotypes together, there is a 1/2 chance one of their children will have the genotype Oo and a 1/2 chance their child will have the genotype of oo. Only a genotype of oo will express purple skin, so there is a 1/2 chance they will have a child expressing purple skin. However, we are asked to find the probability of them having two children who express purple skin, so to find this we multiply the individual probability of each child having purple skin. Therefore we receive and answer of $$ (\frac{1}{2})(\frac{1}{2})=\frac{1}{4} $$, or a 25% chance both children express purple skin.

### Subject: Physics

A ball is thrown from rest at an angle of 30 degrees above the horizontal ground, at a velocity of 5.0 m/s. Find how far the ball will travel before hitting the ground again.

First, you must split the velocity vector up into it's directional components. So, making the x axis in the horizontal direction, and the y axis in the vertical direction, we must find the velocity in the x and y directions using the angle of the trajectory of the ball. The y component will be equal to sin(30)(5.0m/s), or the sine of the angle multiplied by the velocity. $$ V_{y}=sin(30)(5.0m/s)= 2.5m/s $$ . Now the x component of the velocity will be found using the cosine of the angle, $$ V_{x}=cos(30)(5.0m/s)=2.2/s $$. Next, we must find the time that the ball stays in the air because this is the one component of the trajectory that is the same along the x and the y directions. To do this we will use the velocity in the y direction because we know the acceleration in the y direction which is just the acceleration of gravity, or $$ 9.81m/s^2 $$. The acceleration of an object is equal to the change in velocity divided by the change in time, so rearranging this to solve for change in time we get, $$ \frac{\delta V_{y}}{a_{y}}= \delta t $$ . where $$ \delta V_{y} $$ is the change in velocity in the y direction, $$ a_{y} $$ is the acceleration in the y direction, and $$ \delta t $$ is the change in time. Halfway through the trajectory the ball will reach a velocity of 0 m/s, between when the balls stops moving upwards and begins to move downwards or in the negative y direction. So, using this knowledge we can say the the change in velocity in the y direction is 2.5m/s - 0m/s= 2.5m/s. Now since that is only halfway through the trajectory, when we use this value, we must double the time we get. So, plugging these values in, we get $$ \frac{2.5m/s}{9.81m/s^2}= (.26s) $$, and multiplying this value of time by two, (.26s)(2)=.52s. So, now we can use this value of time to calculate the distance traveled in the horizontal x direction using the relationship of $$ \delta x=V_{ix} t + \frac{1}{2} a_{x} t^2 $$ where $$ \delta x $$ is the change in distance the ball travels in the x direction, $$ V_{ix} $$ is the initial velocity in the x direction, $$ t $$ is the time traveled before hitting the ground, and $$ a_{x} $$ is the acceleration in the x direction. The acceleration in the x direction, however is 0, so now plugging in the other values we get, $$ \delta x= (2.2m/s)(5.2s)= 11 m $$. Therefore the ball travels about 11 meters before hitting the ground.