Arocket accelerates upward from rest, due to the first stage, with a constant acceleration of a1 = 85 m/s^2 for t1 = 46 s. the first stage then detaches and the second stage fires, providing a constant acceleration of a2 = 31 m/s^2 for the time interval t2= 41 s. part (a) enter an expression for the rocket's speed, v1, at time t1 in terms of the variables provided.
part (b) enter an expression for the rocket's speed, v2, at the end of the second period of acceleration, in terms of the variables provided in the problem statement.
part (c) using your expressions for speeds v1 and v2, calculate the total distance traveled, in meters, by the rocket from launch until the end of the second period of acceleration.

(a) = 3910

(b) = 5181

(c) =  2762955 m

Explanation:

Using equations of uniformly accelerated motion:

Equations:

(a) Using equation 1:

⇒ Because it upward from rest

(b) Using equation 1:

⇒ Because is the  is the initial speed in the movement before changing acceleration

(c) Using equation 2:

 ⇒ and Because it upward from rest

⇒

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

12000j is work done