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Colt1911 [192]
3 years ago
7

Suppose you want to design an air bag system that can protect the driver at a speed 100 km/h (60 mph) if the car hits a brick wa

ll.
Estimate how fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver?
Physics
1 answer:
34kurt3 years ago
3 0

When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.

Now, given that:

  • the speed of the car which is the initial velocity (u) = 100 km/h before it hits the wall.
  • after hitting the wall, the final velocity will be (v) = 0 km/h

Assumptions:

  • Suppose we make an assumption that the distance travelled during the collision of the car with the brick wall (S) = 1 m
  • That the car's acceleration is also constant.

∴

For a motion under constant acceleration, we can apply the kinematic equation:

\mathsf{v^2 = u^2 + 2as}

where;

v = final velocity

u = initial velocity

a = acceleration

s = distance

From the above equation, making acceleration (a) the subject of the formula:

\mathsf{v^2 - u^2 =2as }

\mathsf{a = \dfrac{v^2 - u^2 }{2s}}

The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.

since 1 km/h = 0.2778 m/s

100 km/h = 27.78 m/s

\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}

\mathsf{a = \dfrac{- 771.7284 }{2}}

a = - 385.86 m/s²

Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration and velocity is;

v = u + at

where;

v = 0

-u = at

\mathsf{t = \dfrac{-u}{a}}

\mathsf{t = \dfrac{-27.78}{-385.86}}

t = 0.07 seconds

An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.

Thus, we can conclude that the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.

Learn more about the kinematic equation here:

brainly.com/question/11298125?referrer=searchResults

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7 0
3 years ago
What type of motion occurs when an object spends around and axis without altering its linear position?
ch4aika [34]
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7 0
3 years ago
Suppose that you lift four boxes individually, each at a constant velocity. The boxes have weights of 3.0 N, 4.0 N, 6.0 N, and 2
Alona [7]

Answer:

The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

Explanation:

Worked : work can be defined as the product of force and distance.

The S.I unit of work is Joules (J).

Mathematically it can be represented as,

W = F×d.................. Equation 1

d = W/F.............................. Equation 2

where W = work, F = force, d = distance.

<em>Given: W = 12 J</em>

(i) for the 3.0 N weight,

using equation 2

d = 12/3

d= 4 m.

(ii) for the 4.0 N weight,

d = 12/4

d = 3 m.

(iii) for the 6.0 N weight,

d = 12/6

d = 2 m.

(iv) for the 2.0 N weight,

d = 12/2

d = 6 m

Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

8 0
3 years ago
Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v
mihalych1998 [28]

Answer:

a.

\displaystyle a(0 )=8.133\ m/s^2

\displaystyle a(2)=2.05\ m/s^2

\displaystyle a(4)=0.52\ m/s^2

b.\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15

c. t=9.9 \ sec

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

\displaystyle V(t)=a(1-e^{bt})

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

\displaystyle a=11.81\ ,\ b=-0.6887

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

\displaystyle V(t)=11.81(1-e^{-0.6887t})

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})

\displaystyle a(t)=8.133547\ e^{-0.6887t}

For t=0

\displaystyle a(0)=8.133547\ e^o

\displaystyle a(0 )=8.133\ m/s^2

For t=2

\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}

\displaystyle a(2)=2.05\ m/s^2

\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}

\displaystyle a(4)=0.52\ m/s^2

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C

\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

\displaystyle x(0)=0=>11.81\times1.45201+C=0

Solving for C

\displaystyle c=-17.1482\approx -17.15

Now we complete the equation for the distance

\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100

Rearranging

\displaystyle t+1.45\ e^{-0.6887t}=9.92

We define an auxiliary function f(t) to help us find the value of t.

\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92

Let's try for t=9 sec

\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92

Now with t=9.9 sec

\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184

That was a real close guess. One more to be sure for t=10 sec

\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).  

At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}

7 0
3 years ago
A piece of copper wire with thin insulation, 200 m long and 1.00 mm in diameter, is wound onto a plastic tube to form a long sol
bearhunter [10]

Answer:

 N= 3

Explanation:

For this exercise we must use Faraday's law

          E = - dФ / dt

         Ф = B . A = B Acos θ

tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives

         E = - A cos θ (B - B₀) / t

The angle enters the magnetic field and the normal to the area is zero

         cos 0 = 1

         A = π r²

   

In the length of the wire there are N turns each with a length L₀ = 2π r

          L = N (2π r)

          r = L / 2π N

    we substitute

          A = L² / (4π N²)

The magnetic field produced by a solenoid is

           B = μ₀ N/L   I

for which

            B₀ = μ₀  N/L   I

           

The final field is zero, because the current is zero

            B = 0

We substitute

           E = - (L² / 4π N²)  (0 - μ₀ N/L I) / t

           E = μ₀ L I / (4π N t)

           N = μ₀ L I / (4π t E)

The electromotive force is E = 0.80 mV = 0.8 10⁻³ V

let's calculate

           N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]

           N  = 320 10⁻⁷ / 9.6 10⁻⁶

           N = 33.3 10⁻¹

          N= 3

           

7 0
3 years ago
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