To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.
From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by
Where,
Final velocity
Initial Velocity
a = Acceleration
x = Displacement
Acceleration can be expressed in terms of the drag coefficient by means of
Frictional Force
Force by Newton's second Law
Where,
m = mass
a= acceleration
Kinetic frictional coefficient
g = Gravity
Equating both equation we have that
Therefore,
Re-arrange to find x,
The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.
Internal.......................
Please see the solution in the attached drawing.
Read it from the top down.
If you ever come back again, with one that has
an AC source and an L or a C in the circuit, I'll
be looking for more than 5 points for that one. :-)
Answer:
The banking angle required is .
Explanation:
Banking of a road is the act of constructing a road along a curved path at a certain angle to avoid skidding-off of vehicles plying it. Centripetal force is required to pull the object moving with a velocity 'v' towards the center of the curve for stability.
The velocity of a car navigating a banked road is given by:
v =(rg ÷ tanθ)
where: r is the radius of the road, g is the gravitational force and θ is the banking angle.
⇒ = rg ÷ tanθ
tanθ =
θ =
= (given that g = 10)
=
= 1.4663
θ =
The banking angle required is .
Start with basic concept that you have the same amount of watts it's just how far away you are dictates the watts/meter squared (per area). As the distance from the object is increased, the area increases as distance squared (think of a sphere of increasing radius, as you increase the radius, the surface area increases as r^2, double radius, you get 4 times area, triple radius, 9 times area, and so forth). so safe from acoustical effects (reflections off walls, etc) as you increase the distance from a speaker, the sound will decrease by a factor of the distance squared, as before 2x distance =1/4 power, 3x distance =1/9 power.
<span>so the only thing left out of your questions is what distance from speaker is the level at 10^-6 W/m^2 and what distance is point B from the speaker, and apply the 1/distance^2 principle. </span>
<span>The power at point B (PB) is the following: </span>
<span>power at given power level (we will call this point A or PA). </span>
<span>PB=PA*(DA/DB)^2 </span>
<span>Where PB is power level at PB (in W/m^2) </span>
<span>PA=power at point A or 10^-6 W/m^2 </span>
<span>DA=distance from speaker to point A </span>
<span>DB=distance from speaker to point B. </span>