A 45-caliber bullet shot straight up from the surface of the moon would reach a height of s = 832t − 2..6t2 ft ater t seconds. On Earth, in the absence of air, its height would be s = 832t − 16t2 ft after t seconds. a. How long will the bullet be aloft on the moon? b. How high will the bullet go on the moon? c. How long will the bullet be aloft on Earth? d. How high will the bullet go on Earth?

A 45-caliber bullet shot straight up from the surface of the moon would reach a height of s = 832t − 2..6t2 ft ater t seconds. On Earth, in the absence of air, its height would be s = 832t − 16t2 ft after t seconds. a. How long will the bullet be aloft on the moon? b. How high will the bullet go on the moon? c. How long will the bullet be aloft on Earth? d. How high will the bullet go on Earth?

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter1: Physics And Measurement

Section: Chapter Questions

Problem 1.29P

See similar textbooks

Similar questions

The radius of Mars is approximately 2,100miles and it's acceleration due to gravity is 0.38g, where g is the acceleration due to gravity in Earth. Determine the velocity of escape of a body if it startto move away from Mars at a distance of one-tenth ofit'sradius aboveit's surface. a. 1.689 mi/sec c. 0.986 mi/sec e. none of the choices b. 0.689 mi/sec d. 1.896 mi/sec
I was able to understand the attached problem and have the answer now for mu. So the equation I have is v=150e-6.0788 x My question, then, is how do we find the TIME to slow the sled to 15 mi/hr ? The answer is supposed to be 35.533 seconds, but I don't know how to get that. Thanks!
A cubic meter (m) of air weighs about 1.25 kg. A decent size classroom has a volume of roughly 800 m, What is the mass of air in such a room? O 1000 kg. 1000g. O 1 kg. O 1 million grams.
1. What is E in the equation 2.1 derived?2. Is E in the eq 2.1 always positive? Why? 3. What is Δx in the derived equation?4. Is it always positive in eq 2.1?
云2 A eclass.yorku.ca U PHYS18... M (2,045)... M(98) - fa.. 105-Fe... Future S. U of Study... Ly Assignm... YouTube M(2,045)... HYS1800 A - Engineering Mechanics (Fall 2021-2022) My courses / SC/PHYS1800 A - Engineering Mechanics (Fall 2021-2022) / Work and Energy / Assignment 6 A car travels at a steady 30.3 m/s around a horizontal curve of radius 111 m. What is the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding? Answer: Next page MacBook Air DD 08 F3 64 F5 F6 F4 F2
10. A diver bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.92 m/s and her takeoff point is 1.994 m above the pool. How long (in seconds) are her feet in the air? (Assume motion is in 1D) 0.353 s 1.966 s 1.153 s 11.3 s 11. How far does light travel in 5 nanoseconds? Express your answer in feet. Assume the travels in a vacuum where its speed is 186,000 miles per second. a. 1 foot b. 5 feet a. b. C. d. C. 29.7 feet d. 186,000 feet
An object is dropped from the top of a cliff 600 meters high. Its height above the ground t seconds after it is dropped is 600−4.9t^2. Determine its speed 9 seconds after it is dropped.
3. The velocity of the body is given by V(t)=sin(xt), where the velocity is given in m/s and t is in seconds. The distance covered in meters between t=1/4 and ½ seconds is close to: A. -0.225 С. 1.2251 В. 0.225 D. 1 4. Evaluate the integral of xy with respect to y and to x, for the limits from x=0 to x=1, and the limits from y=1 to y=2. A. 7/9 C. % B. % D. 5/8
S‏‏at‏e‏l‏‏lite. ‏P‏h‏‏ob‏o‏s‏‏ㅤh‏a‏s‏‏ㅤa‏ㅤ‏r‏‏ad‏i‏u‏‏sㅤ‏o‏f‏‏ㅤ11ㅤ‏k‏m‏‏ㅤw‏h‏i‏‏le‏ㅤ‏M‏‏ar‏s‏ㅤ‏‏ha‏s‏ㅤ‏‏aㅤ‏r‏a‏‏di‏u‏s‏‏ㅤofㅤ3390ㅤkm.ㅤ‏P‏h‏‏ob‏o‏s‏‏ㅤh‏a‏s‏‏ㅤa‏ㅤ‏c‏‏ir‏c‏u‏‏la‏r‏ㅤ‏‏or‏b‏i‏‏tㅤ‏a‏r‏‏ou‏n‏d‏‏ㅤM‏a‏r‏‏sㅤ‏w‏h‏‏ic‏h‏ㅤ‏‏is‏ㅤ‏i‏‏nc‏l‏i‏‏ne‏d‏ㅤ‏‏atㅤ0°‏ㅤ‏t‏‏oㅤ‏M‏a‏‏rs‏’‏s‏‏ㅤe‏q‏u‏‏ator.ㅤ‏T‏h‏‏eㅤ‏p‏e‏‏ri‏o‏d‏‏ㅤo‏f‏ㅤ‏‏Ph‏o‏b‏‏os‏ㅤ‏i‏‏sㅤ7h 40m‏,‏ㅤ‏‏wh‏i‏l‏‏eㅤ‏t‏h‏‏eㅤ‏t‏i‏‏me‏ㅤ‏i‏‏tㅤ‏t‏a‏‏ke‏s‏ㅤ‏‏fo‏r‏ㅤ‏‏Ma‏r‏s‏‏ㅤt‏o‏ㅤ‏‏co‏m‏p‏‏le‏t‏e‏‏ㅤa‏ㅤ‏f‏‏ul‏l‏ㅤ‏‏ro‏t‏a‏‏ti‏o‏n‏‏ㅤa‏r‏o‏‏un‏d‏ㅤ‏‏it‏s‏ㅤ‏‏ax‏i‏s‏‏ㅤisㅤ24h 37m. (Q.W‏here‏ㅤ‏o‏‏nㅤ‏t‏h‏‏eㅤ‏h‏o‏‏ri‏z‏o‏‏nㅤ‏d‏o‏‏es‏ㅤ‏P‏‏ho‏b‏o‏‏sㅤ‏r‏i‏‏se‏?‏ㅤExplainㅤwhy.) Whare does phobos raise is the question plz answer it , I don't need 11.4 hrs answer
I II E B C. F Compare B and C for brightness. A. B>C B. C>B С. В-С
1.67 Ka into ma 3.296X10° Km into mm. 4.967 Kg into da S8961m into (m 6.3.45X lo? m/s into Km/hr 7.234x 63Kmlhre into ms 8.233 um into m 9,432x10" Kq into9
The mass of the sun is 2.0 x 1030 kg. a. What is the mass of the sun in Tg? ball park 1021 Tg b. What is the mass of the sun in ounces?