A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust ends. The rocket has mass 2 kg and thrust force 35 N. Neglecting air resistance, determine (a) the speed of the rocket when the thrust ends, (b) the maximum height reached by the rocket, and (c) the speed of the rocket when it returns to the ground.
Answers
Answer:
a)
b)
c)
Explanation:
a) Let's calculate the work done by the rocket until the thrust ends.
But we know the work is equal to change of kinetic energy, so:
b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.
At the maximum height the velocity is 0, so v(f) = 0.
c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.
Using the same equation of part b)
The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.
I hope it helps you!
Related Questions
There are three homes being built, each with an identical deck on the back. Each deck is comprised of two separate areas. One area is 112.5 square feet, while the other is136,4 square feet. What is the total square footage of the decks for all three homos? Your answer should be to the nearest tonth of a square
A bathroom scale uses four internal strain gauges to measure the displacement of its diaphragm as a means for determining load. Four active gauges are used in a bridge circuit. The gauge factor is 2, and the gauge resistance is 120 Ω for each gauge. If an applied load to the diaphragm causes a tensile strain on R1 and R4 of 40 x 10-6, while gauges R2 and R3 experience a compressive strain of 40 x 10-6, estimate the bridge deflection voltage. The supply voltage is 5 V. a,b,c or d? a.400 x 10-6 V b.600x10-6V c.800x10-6V d.1000 x 10-6 V
IN JAVA,Knapsack ProblemThe file KnapsackData1.txt and KnapsackData2.txt are sample input filesfor the following Knapsack Problem that you will solve.KnapsackData1.txt contains a list of four prospective projects for the upcoming year for a particularcompany:Project0 6 30Project1 3 14Project2 4 16Project3 2 9Each line in the file provides three pieces of information:1) String: The name of the project;2) Integer: The amount of employee labor that will be demanded by the project, measured in work weeks;3) Integer: The net profit that the company can expect from engaging in the project, measured in thousandsof dollars.Your task is to write a program that:1) Prompts the user for the number of work weeks available (integer);2) Prompts the user for the name of the input file (string);3) Prompts the user for the name of the output file (string);4) Reads the available projects from the input file;5) Dolves the corresponding knapsack problem, without repetition of items; and6) Writes to the output file a summary of the results, including the expected profit and a list of the bestprojects for the company to undertake.Here is a sample session with the program:Enter the number of available employee work weeks: 10Enter the name of input file: KnapsackData1.txtEnter the name of output file: Output1.txtNumber of projects = 4DoneFor the above example, here is the output that should be written to Output1.txt:Number of projects available: 4Available employee work weeks: 10Number of projects chosen: 2Number of projectsTotal profit: 46Project0 6 30Project2 4 16The file KnapsackData2.txt, contains one thousand prospective projects. Your program should also be able to handle this larger problem as well. The corresponding output file,WardOutput2.txt, is below.With a thousand prospective projects to consider, it will be impossible for your program to finish in areasonable amount of time if it uses a "brute-force search" that explicitly considers every possiblecombination of projects. You are required to use a dynamic programming approach to this problem.WardOutput2.txt:Number of projects available: 1000Available employee work weeks: 100Number of projects chosen: 66Total profit: 16096Project15 2 236Project73 3 397Project90 2 302Project114 1 139Project117 1 158Project153 3 354Project161 2 344Project181 1 140Project211 1 191Project213 2 268Project214 2 386Project254 1 170Project257 4 427Project274 1 148Project275 1 212Project281 2 414Project290 1 215Project306 2 455Project334 3 339Project346 2 215Project356 3 337Project363 1 159Project377 1 105Project389 1 142Project397 1 321Project399 1 351Project407 3 340Project414 1 266Project431 1 114Project435 3 382Project446 1 139Project452 1 127Project456 1 229Project461 1 319Project478 1 158Project482 2 273Project492 1 142Project525 1 144Project531 1 382Project574 1 170Project594 1 125Project636 2 345Project644 1 169Project668 1 191Project676 1 117Project684 1 143Project689 1 108Project690 1 216Project713 1 367Project724 1 127Project729 2 239Project738 1 252Project779 1 115Project791 1 110Project818 2 434Project820 1 222Project830 1 179Project888 3 381Project934 3 461Project939 3 358Project951 1 165Project959 2 351Project962 1 316Project967 1 191Project984 1 117Project997 1 187
If 100 J of heat is added to a system so that the final temperature of the system is 400 K, what is the change in entropy of the system? a)- 0.25 J/K b)- 2.5 J/K c)- 1 J/K d)- 4 J/K
Answers
Answer:
Compression ratio(r)
Explanation:
Otto cycle:
Otto cycle is an ideal cycle for all working petrol engine.It have four processes in which two are constant volume process and other two are reversible adiabatic or we can say that isentropic processes.All petrol engine works on Otto cycle.
The efficiency of Otto cycle given as follows
Where r is the compression ratio and γ is heat capacity ratio.
So from above we can say that the efficiency of Otto cycle depends onl;y on compression ratio (r).
Answers
Answer:
Time period = 41654.08 s
Explanation:
Given data:
Internal volume is 210 m^3
Rate of air infiltration
length of cracks 62 m
air density = 1.186 kg/m^3
Total rate of air infiltration
total volume of air infiltration
Time period
Answers
Answers
Answer:
Explanation:
The concept of Hooke's law was applied as it relates to deformation.
The detailed steps and appropriate substitution is as shown in the attached file.
Required:
What is the average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G?
Answers
The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.
Why reducing leads to increasing wages?
Reducing such a need to move in between multiple tabs, the split-screen has been valuable for increasing wages. In the several instances running a two or more desktop system will allow different programs to run throughout multiple devices. That works with the same process on both PC and laptop monitors.
Just display them side by side, instead of the switching among both the apps that has been used frequently. In this phase, an app that the snap to either left or right occupies a third of the display, and yet another app holds the two-thirds remaining. It refers to Split-Screen Mode.
Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization.
Therefore, The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.
Learn more about average power on:
#SPJ2
Answers
Answer:
D) AND gate.
Explanation:
Given that:
A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print
These conditions are:
1. The printer's electronic circuits must be energized.
2. Paper must be loaded and ready to advance.
3. The printer must be "on line" with the microprocessor.
Now; if these conditions are met the logic gate produces a HIGH output indicating readiness to print.
The objective here is to determine the basic logic gate used in this circuit.
Now;
For NOR gate;
NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.
For NOT gate.
NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.
Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".
Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.