As a civil engineer for your city, you have been assigned to evaluate the purchase of spring-loaded guard rails to prevent cars from leaving the road. in response to a request for proposals*, one company states their guard rails are perfect for the job. each section of their guard rails consists of two springs, each having a force constant 3.13400 105 n/m with a maximimum distance of compression of 0.614 m. (according to the manufacturer, beyond this compression the spring loses most of its ability to absorb an impact elastically.) the largest vehicle the guardrails are expected to stop are trucks of mass 4550.000 kg. what is the maximum speed at which these guard rails alone can be expected to bring such vehicles to a halt within the stated maximum compression distance? (assume the vehicles can strike the guard rail head on and that the springs are perfectly elastic.)
m/s
given your result which section of road often features such a speed
school zone
large road
highway
guard rails are pointless if the acceleration they create seriously injures passengers. one important safety factor is the acceleration experienced by passengers during a collision. calculate the maximum acceleration of the vehicle during the time in which it is in contact with the guard rail.
m/s^2
if the highway department considers 20 g\'s the maximum safe acceleration, is this guard rail safe in regards to acceleration?

a) v = 7,207 m / s

, b) a = 42.3 m / s²

Explanation:

We will solve this exercise using the concept of mechanical energy, We will write it in two points before the car touches the springs and in point of maximum compression

Initial

    Em₀ = K = ½ m v²

Final  

    = 2 Ke = ½ k x²

The two is placed because each barred has two springs and each does not exert the same force

    Emo =

     ½ m v² = 2 ½ k x²

     v = √(2k/m) x

     v = √ (2 3,134 10⁵/4550) 0.614

     v = 7,207 m / s

Let's take this speed to km / h

     v = 5,096 m / s (1km / 1000m) (3600s / 1h)

     v = 25.9 km / h

This speed is common in school zones

Let's use kinematics to calculate the average acceleration

      vf² = v₀² - 2 a x

       0 = v₀² - 2 a x

       a = v₀² / 2 x

       a = 7,207²/2 0.614

       a = 42.3 m / s²

We buy this acceleration with the acceleration of gravity

       a / g = 42.3 / 9.8

       a / g = 4.3

This acceleration is well below the maximum allowed

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

what experimental evidence led scientists to change from the previous model to this one?

a few of the positive particles aimed at a gold foil seemed to bounce back.

the colors of light emitted from heated atoms had very specific energies.

experiments with water vapor showed that elements combine in specific proportions.

cathode rays were bent in the same way whenever a magnet was brought near them.