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Lelu [443]
3 years ago
12

A 72.5-kg person is riding in a car moving at 20.5 m/s (assume this is the positive direction) when the car runs into a bridge a

butment. This problem will illustrate why the invention of the airbag dramatically improved the safety of automobiles. show answer Incorrect Answer 50% Part (a) Calculate the horizontal component of the average force, in newtons, on the person if he is stopped by a padded dashboard that compresses an average of 1.15 cm.Calculate the horizontal component of the average force, in newtons, on the person if he is stopped by an air bag that compresses an average of 13.5 cm
Physics
2 answers:
MAXImum [283]3 years ago
6 0

Explanation:

Given that,

Mass of the person, m = 72.5 kg

Initial speed of the car, u = 20.5 m/s

(a) Let F_x is the horizontal component of the average force, in newtons, on the person if he is stopped by a padded dashboard that compresses an average of 1.15 cm, x = 0.0115 m and v = 0

Using the second law of motion to find it as:

F_x=m\times a

Using the third equation of kinematics to find a.

F_x=m\times \dfrac{v^2-u^2}{2s}

F_x=-m\times \dfrac{u^2}{2s}

F_x=-72.5\times \dfrac{(20.5)^2}{2\times 0.0115}

F_x=-1.32\times 10^6\ N

(b) Here, x = 13.5 cm = 0.135 m

F_x=-m\times \dfrac{u^2}{2s}

F_x=-72.5\times \dfrac{(20.5)^2}{2\times 0.135}

F_x=-1.12\times 10^5\ N

Hence, this is the required solution.

uranmaximum [27]3 years ago
6 0

Answer:

A)  f_{dashboard} = 1324701 N

B)  f_{airbag} = 112844.90 N

Explanation:

Given data:

mass of person = 72.5 kg

velocity of person = 20.5 m/s

we know that from newton's 2nd law

f = ma

f = m \frac{v^2 - u^2}{2s}

final velocity is v = 0

f = - \frac{mu^2}{2s}

f = \frac{72.5\times 20.5^2}{2\times 1.15\times 10^{-2}}

f = 1324701 N

B)

f_{airbag} = f = \frac{72.5\times 20.5^2}{2\times 1.3.5\times 10^{-2}}

f_{airbag} = 112844.90 N

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3 years ago
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A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force
Tom [10]

Answer:

VB -  VA  =  - 33.4

Explanation:

Generally the workdone in moving the proton is mathematically represented as

     W  =  KE_f  - KE_i

Where KE_i \ and \  KE_f \  are\  the\  initial  \  and  \  final \  kinetic \  energy

So

    KE_i  =  \frac{1}{2} m v_a^2

Here v_a is the velocity at A with value  50 m/s

So

    KE_i  =  \frac{1}{2} (1.67*10^{-27}) * 50^2

    KE_i  = 2.09 *10^{-24} \  J

Also  

     KE_f  =  \frac{1}{2} m v_b^2

Here v_a is the velocity at A with value 80 km/s = 80000 m/s

=>   KE_f  =  \frac{1}{2} (1.67*10^{-27}) * 80000^2

=>   KE_f  = 5.34 *10^{-18} \  J

 So

    W  =   5.34 *10^{-18}  - 2.09 *10^{-24}

     W  =   5.34 *10^{-18}  m/s

Now this workdone is also mathematically represented as

     W =  q *  V

So  

    q *  V =   5.34 *10^{-18}

Here  q =  1.60*10^{-19} C

So

        V =   \frac{5.34 *10^{-18} }{1.60*10^{-19}}

         V =   33.4 \  V

Generally proton movement is in the direction of the electric field it means that  VA>VB

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    VB -  VA  =  - 33.4

8 0
3 years ago
A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w
cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

\displaystyle P=\frac{W}{t}

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

W=F.x

The second Newton's law gives us the net force as

F=m.a

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

v_f=v_o+at

v_f=at

Solving for a

\displaystyle a=\frac{v_f}{t}={5.4}{1}

a=5.4\ m/s^2

The distance traveled in the interval is given by

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Since vo=0

\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}

x=2.7\ m

The force is given by

F=m.a

We don't know the value of m, so the force is

F=2.7m

Computing the work done by the sprinter

W=F.x=2.7m(5.4)

W=14.58m

The power is finally computed

\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}

P=14.58m

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}

a=8.8\ m/s^2

The distance is

\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}

x=3.8\ m

The net force is

F=m(8.8)=8.8m

The work done by the sprinter is now computed as

W=8.8m(3.8)=33.44m

At last, the output power is

\displaystyle P=\frac{33.44m}{0.5}=66.88m

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0
3 years ago
If the torque required to loosen a nut that holds a wheel on a car has a magnitude of 55 n·m, what force must be exerted at the
erastova [34]

Either 175 N or 157 N depending upon how the value of 48° was measured from.    
You didn't mention if the angle of 48° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.    
Since we need a torque of 55 N·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 N·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.    
Ideally, the force will have a 0° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".     
If the angle of 48° is from the normal to the lug wrench, the usefully applied power will be:  
U = F*cos(48)  
where  
U = Useful force  
F = Force applied    
So solving for F and calculating gives:  
U = F*cos(48)  
U/cos(48) = F  
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174.8537563 N = F    
So 175 Newtons of force is required in this situation.    
If the 48° is from the lug wrench itself, that means that the force is 90° - 48° = 42° from the normal. So doing the calculation again (this time from where we started plugging in values) we get  
U/cos(42) = F  
117/0.743144825 = F  
157.4390294 = F    
Or 157 Newtons is required for this case.
6 0
3 years ago
A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853
Norma-Jean [14]

Recall that

{v_f}^2-{v_i}^2=2a\Delta x

where v_i and v_f are the initial and final velocities, respecitvely; a is the acceleration; and \Delta x is the change in position.

So we have

\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)

\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

7 0
3 years ago
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