Acar traveling at 43 km/h hits a bridge abutment. a passenger in the car moves forward a distance of 54 cm (with respect to the road) while being brought to rest by an inflated air bag. what magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of 49 kg?

F = -6472.9 N

F= -6.47 kN

Explanation:

First of all you have to convert the data to SI  units

so for the velocity you have :

Vi = 43km/h *(1000m/1km)*(1h/3600s)  ---> using conversion factors

Vi= 11.9444 m/s

dX : distance the passanger moves

dX = 54cm*(1m/100cm)   -->  using conversion factors

dX = 0.54 m

Now to calculate the force we are going to use the sum of focers equals to mass for acceleration:

Sum F = m*a

We have to find a so we are going to use the velocity's formula as follows to solve a:

Vf ^2 = Vi^2 +2*a*dX

Vf=0  --> the passenger does not move after the airbag inflates.

a= -(Vi^2)/(2*dX)

you solve de acceleration with the data you hae and you will find

a = -132.1 m/ s^2

Now you can solve the Sum F equation

Sum F = 49 Kg * (-132.1 m/s^2)

F = -6472.9 N

F= -6.47 kN

  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

hypothesising or postulating