Tutor profile: Benjamin B.
Subject: Physics (Thermodynamics)
An air conditioner is placed inside an adiabatic, well insulated room. The air conditioner is connected to an outlet and turned on. Describe what happens to the temperature of the room using the first law of thermodynamics as your reasoning.
First and foremost, the first law of thermodynamics says that energy cannot be created or destroyed, only transferred. This is the case with this air conditioning problem. Many people would think that the temperature of the room would decrease, it is an air conditioner, right? The problem here is that the room is adiabatic, and doesn't let heat transfer in or out of the room. In practical use of air conditioners, the outside air is used as thermal reservoir to disperse the heat. Because energy is never created or destroyed, an air conditioner has to transfer heat (energy) from the air to somewhere else to cool down a room. However, in our case, there is no thermal reservoir. There is only a well insulated room, meaning the electricity (energy) enters and cannot escape, thus increasing the temperature of the room.
Natalie decides to walk to her mailbox. Her velocity is given by the equation v=-2(t-4), where (v) is velocity in feet per second and (t) is time in seconds. A positive velocity indicates walking away from her home, a negative indicates returning back. Assume she only walks on a straight-line path. What is the total distance she travels between leaving the house at time (t)=0 and returning from the mailbox?
The first step is to define and write down what we know. time (seconds), t=0 velocity (feet/second)= -2(t-4) We are trying to find distance traveled. One of the most important things to know in calculus is how distance, velocity and acceleration relate. For this problem it is important to know that distance is the integral of velocity. This indicates we need to take an integral of our velocity equation. Taking an integral from time (t)=0 to (t)=(t), time, we find the following: distance=-t^2+8t. This leaves a problem, however, we do not know the time. We can find this from the original velocity equation. This is the tricky part, we actually want to solve for the time when velocity is zero, which corresponds to the furthest distance from the house. This may not always be the case, but we can see from the the velocity equation it is not possible to be any farther from the house than when her velocity is at 0. At velocity=0, Natalie spends the rest of her run returning to the house. We solve for 0=-2(t-4) to get the time where the distance is maximum (tmax=4). Next we plug 4 into our distance formula to get the maximum distance from the house, dmax=16ft. We know that Natalie returns this distance in a straight line and follows the equation -2(t-4), thereby walking the same distance back. Natalie therefore travels 32 feet to and from her mailbox. Extra: Keep in mind this could have been solved by figuring out the area under the two triangles of the equation v=-2(t-4) from 0 to 8 seconds.
If a ball is shot out of a cannon resting on the floor (height=0), write an expression for the relation between velocity (v) and angle in degrees (p) from the floor to the cannon and the gravitational constant (g) required to have the ball initially hit the ground 100 feet away. Ignore air resistance.
The first step is to write out the information given to us (don't forget units!) and the information we would like to find. We are given the following: velocity, v angle(degrees) ,p initial height=0 final height=0 distance to target (feet), d=100 gravitational constant, g We know we are being asked to write an expression relating velocity, angle and gravitational constant g. Don't be intimidated, working with variables can be straight forward and uses the same steps as solving a regular problem with real numbers! With problems that ask for the relationship between two variables, one will often have to find two or more equations and figure out how to relate them. If we imagine ourselves throwing a ball, we know that how fast ones throws and the angle one throws will determine where the ball lands. We are essentially being asked to find this relationship. The first step is to write an equation for both the vertical and horizontal velocities. Vertical is given by: -0.5(g)(t^2)+(v)sin(p)(t)=height Horizontal is given by: (v)cos(p)(t)=distance Important notes here: These equations come from the general equations of motion. -(g) is the acceleration and is downward, (v)sin(p) is the portion of the initial velocity from the cannon vertically. The initial height is zero. (v)cos(p) is the initial velocity from the cannon horizontally, which increases with time. We know information about when the ball hits the ground after being shot, distance=100ft and height is zero. This means: (v)cos(p)(t)=100ft and -0.5(g)(t^2)+(v)sin(p)(t)=0. There seems to be an extra variable, (t), however. Remember we want (v),(p) and (g) in our answer. We also know that the time is the same at when it lands, this should give us a good indicator that t is the variable we will use to relate (v),(p) and (g). Let us then solve for t in both equations. The horizontal equation becomes t=100/((v)cos(p)(t)) by dividing both sides of the original equation by (v)cos(p)(t) The vertical equation requires the quadratic equation to isolate (-b +/- sqrt(b^2-4ac))/2a. In this case, b=(v)sin(p), a=-0.5g, and c=0. This results in t=(-(v)sin(p) +/- (v)sin(p))/(-g). Since we know that t=0 only at launch, choosing the negative of the +/- is what we want. This results in t=2(v)sin(p)/g and t=100/((v)cos(p)) at the point the ball hits the ground. We then set the two equations equal to each other and solve. By multiplying both sides by (v)cos(p), we are left with 100=(2(v^2)sin(p)cos(p))/g as our final answer. Additional: If we follow where the 100 came from, we know this is the distance from the cannon we want the ball to land. This means that we can change the distance (keeping all else the same) and be able to find the velocity and angle required.
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