A driver of a car hits its brakes and begins to decelerate. If the velocity changes from 60 m/ s to 20/m in 6 s, what is the deceleration of the car?

-6.67 m/s²

Explanation:

Given that,

Initial velocity of car (Vi) = 60 m/s

Final velocity of car (Vf) = 20 m/s

Time taken (T) = 6 seconds

Deceleration (A) = ?

Deceleration is the rate of change of velocity per unit time as the car prepares to rest (slows down)

i.e Deceleration = (Vf - Vi)/T

A = (20 m/s - 60km/s) / 6 seconds

A = -40 m/s / 6 seconds

A = -6.67 m/s² (The negative sign indicates that the car is slowing down)

Thus, the deceleration of the car is -6.67m/s²

  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

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards