Answer:
ΔF=125.22 %
Explanation:
We know that drag force on the car given as
=Drag coefficient
A=Projected area
v=Velocity
ρ=Density
All other quantity are constant so we can say that drag force and velocity can be given as
Now by putting the values
Percentage Change in the drag force
ΔF=125.22 %
Therefore force will increase by 125.22 %.
When distance<span> is increased the amount of </span>force<span> needed will depend on the </span>mass<span> of the object. </span>
Answer:
No photo or graph is there
Explanation:
The equations are analogous to that for linear movement:
acceleration = (final velocity - initial velocity) / time
acceleration = (3000 rpm - 0 rpm) / 2.0 s
a) acceleration = 1500 rpm/s or 25 rp(s^2)
For the displacement
displacement = initial velocity*time + 0.5*acceleration*time^2
displacement = (0)*(2 s) + (0.5)(25 rps^2)*(2 s)^2
b) displacement = 50 revolutions
