Answer:
Explanation:
When we accelerate in a car on a straight path we tend to lean backward because our lower body part which is directly in contact with the seat of the car gets accelerated along with it but the upper the upper body experiences this force later on due to its own inertia. This force is accordance with Newton's second law of motion and is proportional to the rate of change of momentum of the upper body part.
Conversely we lean forward while the speed decreases and the same phenomenon happens in the opposite direction.
While changing direction in car the upper body remains in its position due to inertia but the lower body being firmly in contact with the car gets along in the direction of the car, seems that it makes the upper body lean in the opposite direction of the turn.
On abrupt change in the state of motion the force experienced is also intense in accordance with the Newton's second law of motion.
Efficiency = (useful output) / (input)
Efficiency = (35 J) / (125 J) = 0.28 = 28%
Answer:
1.125m/s^2
Explanation:
Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically
v^2= u^2+2as
Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.
a = ?
u = 0m/s
v = 15m/s
s = 100m
Substituting the values into the formula above
v^2= u^2+2as
15^2=0^2+2×a×100
225= 0+200a
225= 200a
Divide both sides by 200
225/200 = 200a/200
a= 1.125m/s^2
Hence the acceleration of the car is 1.125m/s^2.
Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s
The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N.
<span>Fx = [(233 + 840)/g]*v²/7.5 </span>
<span>v = 32.3*2*π*7.5/60 m/s = 25.37 m/s </span>
<span>The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º </span>
<span>Thh horizontal and vertical forces must balance each other. First the vertical components: </span>
<span>233 + 840 = Ti*cos40º </span>
<span>solve for Ti. (This is the answer to the part b) </span>
<span>Horizontally </span>
<span>[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º </span>
<span>Solve for Th </span>
<span>Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º </span>
<span>using v and Ti computed above.</span>