ES FUILL SCREIN PRINTER VERSION BACK NE Chapter 02, Problem 16 6Your answer is p

Question

ES FUILL SCREIN PRINTER VERSION BACK NE Chapter 02, Problem 16 6Your answer is partially correct. Try again. Over a time interval of 1.51 years, the velocity of a planet orbiting a distant star reverses direction, changing from +23.1 km/s to -23.0 km/s. Find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s2) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration. 28 29 (a) Number 46.1 Units mVs (b) Number 30.53 Units mys 2

Explanation / Answer

a)

Chnage in velocity= - 23.0-23.1= -

46100 m/s

b)

aaceleration , a= change in velocity/time= - 46.1*1000/(1.51*365* 86400)= - 9.681*10^-4 m/s^2

=======

Comment in case any doubt . good luck

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.