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kirza4
10 days ago
5

In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce

leration of 50g that lasts for less than 30 ms, but in a crash with a 50g acceleration lasting longer than 30 ms, a driver is unlikely to survive. Imagine a collision in which a driver's head experienced a 50g accelerationWhat is the highest speed that the car could have had such that the driver survived? Express your answer with the appropriate units.
Chemistry
1 answer:
KiRa [971]10 days ago
8 0

Answer:

Initial velocity, u = 14.7 m/s

Explanation:

It is noted that a driver may endure an acceleration of 50 g for a duration of less than 30 ms, but during a crash where the 50 g acceleration persists beyond 30 ms, survival becomes improbable for the driver.

Let v represent the maximum speed attainable by the car that would allow the driver to survive. We apply a = -50 g and t = 30 ms.

Utilizing the first kinematic equation as follows:

v=u+at

In the case of a crash, the driver's final speed is v = 0

0=u+at

-u=at

-u=-50\times 9.8\times 30\times 10^{-3}

The initial velocity, u = 14.7 m/s

Therefore, the maximum speed the vehicle could reach to ensure the driver's survival is 14.7 m/s, marking this as the solution.

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Alekssandra [968]
Utilize the ideal gas law:

n = PV / RT

P = 100kPa = 100 x 1000 x (9.8 x 10^{-6}) = 0.98 atm
Convert kPa to atm, where 1 Pa = 9.8 x 10^{-6} atm.
T = 293 K
V = 6.8 L
R = 1/12
Substituting all values leads to:
n = 0.272
4 0
9 days ago
Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t
lorasvet [956]

Respuesta:

Un avión fabricado con aluminio puede transportar una mayor cantidad de pasajeros comparado con uno de acero.

Explicación:

La masa total que el avión es capaz de levantar es:

m_{tot}=m_{fuselage}+m_{passangers}

Para el aluminio:

m_{tot}=m_{fus-Al}+m_{pas-Al}

m_{fus-Al}=\delta _{Al}*V_{fuselage}

y

V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]

donde:

  • L es longitud
  • D es diámetro
  • e es grosor

m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}

Para el acero (mismo procedimiento):

m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel

Sabiendo que la masa total que el avión puede levantar es constante y que el aluminio tiene una densidad menor que la del acero, podemos afirmar que el avión de aluminio puede levantar un mayor número de pasajeros.

También es posible estimar un peso promedio de los pasajeros para calcular cuántos podría soportar.

5 0
14 days ago
Solving applied density problems Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across Inter
KiRa [971]

Answer:

thickness is 0.29 cm

Explanation:

To create a fake iron ball out of gold, we must ensure that its mass matches that of the iron ball. Therefore, we first find the volume of the iron ball using the provided diameter, applying the formula of 4/3 pi r^3.

Given the diameter d = 6 cm; thus, the radius r = 3 cm (d/2).

We calculate the volume of the iron ball: 4/3 * 3.14 * 3^3 = 113.04 cm^3.

The corresponding mass of the iron ball is the volume multiplied by its density: 113.04 * 5.15 g/cm^3 = 582.156 g.

This value represents the mass for the gold ball; now we determine the volume of the gold ball using its density.

Volume of gold ball = mass of gold ball/density of gold = 582.156 g/19.3 g/cm^3 = 30.1635 cm^3.

So this volume must correspond to a hollow sphere with an outer radius R = 3 cm and an unknown inner radius r.

Volume of the hollow ball can be represented as: 4/3 pi [R^3 - r^3].

Thus, 30.1635 cm^3 = 4/3 pi [3^3 - r^3].

30.1635 * 3/(4 * 3.14) = 27 - r^3.

Simplifying gives 7.2046 = 27 - r^3, resulting in r^3 = 19.7954.

Therefore, r = 2.7051 cm.

This indicates the thickness is the outer radius minus the inner radius: 3 - 2.7051 = 0.2949 cm.

Rounding to two significant figures yields

the thickness = 0.29 cm.

8 0
6 days ago
In how many grams of water should 25.31 g of potassium nitrate (kno3) be dissolved to prepare a 0.1982 m solution?
lions [985]

Solution:

Molality measures the concentration of a solute in a solution, defined by the amount of solute per specific mass of solvent.

Thus,

Molality = moles of solute / kg of solvent.

Therefore, kg of solvent = moles of solute / molality.

moles of solute = mass / molar mass

= 25.31 g / 101.1 g/mole

= 0.2503 mole.

kg of solvent = 0.2503 mole / 0.1982 m

= 1.263 kg

= 1263 g.

This is the final answer.

6 0
5 days ago
Read 2 more answers
The recommended daily intake of potassium ( K ) is 4.725 g . The average raisin contains 3.513 mg K . Fill in the denominators o
KiRa [971]

Explanation:

It is established that 1 gram is equivalent to 1000 milligrams. We can express this mathematically in the following way.

             \frac{1 g}{1000 mg} or \frac{1000 mg}{1 g}

Thus, to convert grams to milligrams, we simply multiply the number by 1000. Conversely, for converting milligrams back to grams, we divide by 1000.

4 0
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