A rocket blasts off and moves straight upward from the launch pad with constant acceleration. After 2.7 s the rocket is at a height of 93 m. What are the magnitude and direction of the rocket's acceleration?
What is its speed at this elevation?

The magnitude and direction of the rocket acceleration is 68.89 m/s² upward.

The speed of the rocket at the given elevation is 186 m/s.

Explanation:

Given;

time to reach the given height, t = 2.7 s

height reached, h = 93 m

initial velocity of the rocket, u = 0

The magnitude and direction of the rocket acceleration is calculated as;

h = ut + ¹/₂at²

h = 0 + ¹/₂at²

h = ¹/₂at²

a = 2h / t²

a = (2 x 93) / 2.7

a = 68.89 m/s²

the direction of the acceleration is upward.

The speed at this elevation, V = u + at

V = at

V = 68.89 x 2.7

V = 186 m/s

  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