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1 year ago

11

Airplane flight recorders must be able to survive catastrophic crashes. Therefore, they are typically encased in crash-resistant

steel or titanium boxes that are subjected to rigorous testing. One of the tests is an impact shock test, in which the box must survive being thrown at high speeds against a barrier. A 52-kg box is thrown at a speed of 400 m/s and is brought to a halt in a collision that lasts for a time of 6.5 ms. What is the magnitude of the average net force that acts on the box during the collision

1 answer:

chubhunter [2.5K]1 year ago

8 0

Answer:

3.2 MN

Explanation:

Given that

Mass of the box, m = 52 kg

Initial velocity of the box, u = 400 m/s

Final velocity of the box, v = 0 m/s

Time taken for the collision, t = 0.0065 s

Using the equation of motion

V = u + at, we turn around and make acceleration, a the subject of the formula. Now we have,

a = (v - u) / t

a = (0 - 400) / 0.0065

a = 61538.5 m/s²

The acceleration(which is negative acceleration or retar.dation actually) is 61538.5 m/s². We then proceed to is this acceleration in the basic Force equation, to get the magnitude of force needed.

Remember,

Force = mass * acceleration

F = ma, we already have our mass and acceleration, all we do is multiply

F = 52 * 61538.5

F = 3200002 N or 3.2 MN

Therefore, the magnitude of the force that acts on the box during collision is 3.2 MN

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The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em> $1.83\times 10^{7}\,kg$ <em>. The wheels of the locomotive push back on the tracks with a constant net force of </em> $7.50\times 10^{5}\,N$ <em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force ( $F$ ), measured in newtons, then acceleration ( $a$ ), measured in meters per square second, must be constant and can be found by the following expression:

$a = \frac{F}{m}$ (1)

Where $m$ is the mass of the freight train, measured in kilograms.

If we know that $F = 7.50\times 10^{5}\,N$ and $m = 1.83\times 10^{7}\,kg$ , then the acceleration experimented by the train is:

$a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}$

$a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$

Now, the time taken to accelerate the freight train from rest ( $t$ ), measured in seconds, is determined by the following formula:

$t = \frac{v-v_{o}}{a}$ (2)

Where:

$v$ - Final speed of the train, measured in meters per second.

$v_{o}$ - Initial speed of the train, measured in meters per second.

If we know that $a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$ , $v_{o} = 0\,\frac{m}{s}$ and $v = 22.222\,\frac{m}{s}$ , the time taken by the freight train is:

$t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s} }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }$

$t = 542.265\,s$

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

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1 year ago

A stunt car driver testing the use of air bags drives a car at a constant speed of25 m/s for a total of 100m. He applies his bra

PIT_PIT [208]

Answer:

The graphs are attached

Explanation:

We are told that he starts with a constant speed of 25 m/s for a distance of 100 m.

At constant velocity, v = distance/time

time(t) = distance(d)/velocity(v)

t1 = 100/25

t1 = 4 s

Now, we are told that he applies his brakes and accelerates uniformly to a stop just as he reaches a wall 50m away.

It means, he decelerate and final velocity is zero.

Thus;

v² = u² + 2as

0² = 25² + 2a(50)

25² = - 100a

625 = - 100a

a = - 625/100

a = - 6.25 m/s²

v = u + at

0 = 25 + (-6.25t)

25 = 6.25t

t = 25/6.25

t = 4 s

With the values gotten, kindly find attached the distance-time and velocity-time graphs.

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2 years ago

Ancient Greek philosophers spent lots of time thinking about science and imaging explanations for the natural world. What part o

Illusion [34]

Answer:

Testing

Explanation:

Ancient Greek philosophers lived with the ideology to simply contemplate life. This means that their whole life revolved around thinking and questioning everything. This would include creative thinking, because they would sometimes come up with theories which require creativeness. They would often debate with their friends as to why their theory should be accepted or what their opinions were on the matter. More often than not, they argued a lot, and many philosophers went against some powerful people in the community and some were even sentenced to death.

The main process they didn't/couldn't do was the testing. They could never test certain theories because they did not have the means to.

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1 year ago

A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the t

Mice21 [21]

Answer:

The magnitude of the acceleration of the car is 35.53 m/s²

Explanation:

Given;

acceleration of the truck, $a_t$ = 12.7 m/s²

mass of the truck, $m_t$ = 2490 kg

mass of the car, $m_c$ = 890 kg

let the acceleration of the car at the moment they collided = $a_c$

Apply Newton's third law of motion;

Magnitude of force exerted by the truck = Magnitude of force exerted by the car.

The force exerted by the car occurs in the opposite direction.

$F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2$

Therefore, the magnitude of the acceleration of the car is 35.53 m/s²

3 0

2 years ago

A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity

Salsk061 [2.6K]

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

$\omega_i = 2\pi f_1$

$\omega_1 = 2\pi(\frac{610}{60})$

$\omega_1 = 63.88 rad/s$

similarly final angular speed is given as

$\omega_f = 2\pi f_2$

$\omega_2 = 2\pi(\frac{837}{60})$

$\omega_2 = 87.65 rad/s$

angular acceleration of the turbine is given as

$\alpha = 5.9 rad/s^2$

now we have to find the angular displacement is given as

$\theta = \omega t + \frac{1}{2}\alpha t^2$

$\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)$

$\theta = 234.62 radian$

3 0

1 year ago