• 100% Satisfaction Guarantee
GLENN, Teacher
Category: Homework
Satisfied Customers: 633
Experience:  Math Teacher for at least 15 years
79514296
GLENN is online now

# I am looking with my HomeWork Level Physics

### Customer Question

Hi yes I am looking for help with my HomeWork for College Level Physics
Submitted: 1 year ago.
Category: Homework
Customer: replied 1 year ago.
This is the homework I am Having Trouble With
Customer: replied 1 year ago.
Posted by JustAnswer at customer's request) Hello. I would like to request the following Expert Service(s) from you: Help via Email or Text Message. Let me know if you need more information, or send me the service offer(s) so we can proceed.
Customer: replied 1 year ago.
Show all work. Draw diagrams and explain answers as needed. Partial credit is available.1) Falling objects subject to aerodynamic drag accelerate initially but then reach what is called a terminal velocity. The motion from this point on is at constant velocity. Consider the motion of an object that has reached a terminal velocity of 28.6 m/s (down) at a height of 230m. If x=0 (the origin) is at ground level and “up” is the positive direction:a) Write out xf =xi + vit for this motion leaving x and t as variables. The data given represents the t = 0 motion state.
b) Use the equation formulated in part (a) to determine the t value at which the object is 50 m high, and the t value at which the object hits the ground.2) Three students formulate equations to solve the following problem. A lunar lander is making its descent to Moon Base I. The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is 5.0 m above the surface and has a downward speed of 0.8 m/s. With the engine off, the lander is in free fall.Each uses a different coordinate system to describe the motion depicted in the problem statement and formulates both a position, y, and a velocity v, equation.Student 1: y = (0.8 m/s)t + 1/2gt2, v = (0.8 m/s) + gt
Student 2: y = 5.0 m – (0.8 m/s)t – 1/2gt2, v = -(0.8 m/s) – gt
Student 3: y = – (0.8 m/s)t – 1/2gt2, v = -(0.8 m/s) – gt
The equations are based on the general forms:y = yi + vit+ 1/2at2 and
v = vi + at . And all three formulations are “correct”. In each equation g=+1.6 m/s2 and the initial conditions (t=0) describe the lander 5.0 m from the surface moving downward at 0.8 m/s.a) In formulating his or her two equations each student had to choose a real world position of the origin (the position of y=0) relative to the lunar surface, and a real world direction at which the y axis points (either up or down). Draw a diagram for each formulation showing the position of the origin relative to the lunar surface and the real world direction at which the y axis points. Briefly explain your reasoning in each case.b) Each student solves for the t value at which the lander reaches the lunar surface by setting y equal to the position of the lunar surface in his or her chosen coordinate system. What value of y does each student use?c) Use student 3’s formulation and solve for the t and v values of the lander just as it reaches the lunar surface. That is, solve the problem with student #3’s formulation.3. You have a job working for a University research group investigating ozone depletion in the atmosphere. The plan is to collect data on the chemical composition of the atmosphere as a function of the distance from the ground using a mass spectrometer located in the nose cone of a rocket fired vertically. To make sure the delicate instruments survive the launch, your task is to determine the acceleration of the rocket before it uses up its fuel. The rocket is launched straight up with a constant acceleration until the fuel is gone 30 seconds later. To collect enough data, the total flight time must be 5.0 minutes before the rocket crashes to the ground.
Expert:  GLENN replied 1 year ago.

Hi

Thank you for using this site.

How many days do you need the answers for these?