When a 1984 alfa romeo spider sports car accelerates at the maximum possible rate, its motion during the first 20 is extremely well modeled by the simple equation v2x=2pmt where p=3.6104w is the car's power output, m = 1200kg is its mass, and vx is in m/s. that is, the square of the car's velocity increases linearly with time.

a) what is the car's speed at t=11 s ?

b) what is the car's speed at t=18 s ?

c)find a symbolic expression, in terms of p, m, and t, for the car's acceleration at time t.

d)evaluate the acceleration at t=1.0 s .

e) evaluate the acceleration at t=10 s .

a-)

b-)

c-)

d-)

e-)

Explanation:

First, lay out the given data:

P = 3.6 x 10^4 W

m=1200 kg

a)What is the car's speed at t=11 s ?

b) What is the car's speed at t=18 s ?

c) Find a symbolic expression, in terms of P, m, and t, for the car's acceleration at time t.

An expression for the acceleration can be defined by finding the derivative of the velocity equation:

d) Evaluate the acceleration at t=1.0 s

e) Evaluate the acceleration at t=10 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