In,an,emergency,stop,to,avoid,an,accident,a,shoulderstrap,seat,belt,holds,a,60kg,passenger,firmly,in,place.If,the,car,were,initially,traveling,at,90km/h,andcame,to,a,complete,stop,in5.5s,along,a,straight,level,road,what,was,the,average,force,applied,to,the,passenger,by,the,seatbelt?

Now you are going to have to start doing these yourself. It is all force = mass * acceleration

To calculate the average force applied to the passenger by the seatbelt, we can use Newton's second law of motion, which states that force (F) equals mass (m) multiplied by acceleration (a).

1. First, convert the speed from km/h to m/s.
- 90 km/h = 90 * (1000/3600) m/s ≈ 25 m/s

2. Next, identify the acceleration (a). We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Since the car comes to a complete stop, the final velocity (v) is 0 m/s, and the initial velocity (u) is 25 m/s.
- v = u + at
- 0 = 25 + a * 5.5

3. Rearrange the equation to solve for acceleration (a).
- a * 5.5 = -25
- a ≈ -25 / 5.5
- a ≈ -4.55 m/s^2

Note: The negative sign indicates deceleration.

4. Now that we have the acceleration, we can calculate the force applied to the passenger.
- F = m * a
- F = 60 kg * (-4.55 m/s^2)
- The mass of the passenger is given as 60 kg.

5. Calculate the average force.
- F ≈ -273 N

Therefore, the average force applied to the passenger by the seatbelt is approximately 273 Newtons (N). The negative sign indicates that the force is in the opposite direction of the motion (deceleration).