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AleksandrR [38]
1 year ago
12

Two cars, with the same mass and traveling at the same speed, hit large trees head-on. One car has a rigid body that undergoes l

ittle or no deformation in the collision. The other has ``crumple zones'': portions of the body designed to crumple and deform in such a collision. How does this improve the chances that the driver of the second car will survive the event
Physics
1 answer:
nordsb [41]1 year ago
8 0

The crumple zones in the second car will improve the chance of survival of the driver because it will act as shock absorber, reducing the impact of the force on the driver.

<h3>Newton's third law of motion</h3>

According to Newton's third law of motion, action and reaction are equal and opposite.

The car with rigid body will exert maximum force to the driver while the car with crumple zone will exert lesser force to the driver since the crumple zone will act as shock absorber, reducing the impact of the force on the driver.

Thus, the crumple zones in the second car will improve the chance of survival of the driver because it will act as shock absorber, reducing the impact of the force on the driver.

Learn more about Newton's third law of motion here: brainly.com/question/25998091

#SPJ1

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A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police
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Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. x=x_{0}+vt

2. x=x_{0}+v_{0}t+\frac{at^{2}}{2}

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

vt = v_{0}t+\frac{at^2}{2}\\\\   16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}

We simplify the fraction present and rearrange for our formula so that it equals 0:

0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0

In the very last step we factored a common factor t. There is two possible solutions to the equation at t=0 and:

0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\  0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at t=0 s (when the SUV passed the police car) and t=15.56s(when the police car catches up to the SUV)

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Answer:

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Explanation:

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The flux \phi_E can now be determined by using the expression

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\phi_E = 1.52*10^6 Nm^2/C

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3 years ago
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