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Anon25 [30]
3 years ago
6

During a car collision, the knee, thighbone, and hip can sustain a force no greater than 4000 N/ Forces that exceed this amount

could cause dislocations or fractures. Assume that in a collision a knee stops when it hits the car's dashboard. Also assume that the mass of the body parts stopped by the knee is about 20% of the total body mass. a) What is the minimum stopping time interval in needed to avoid injury to the knee if the person is initially traveling at 15 m/s (34 mi/h)? b) what is the minimum stopping distance?
Physics
1 answer:
atroni [7]3 years ago
8 0

Complete Question:

During a car collision, the knee, thighbone, and hip can sustain a force no greater than 4000 N/ Forces that exceed this amount could cause dislocations or fractures. Assume that in a collision a knee stops when it hits the car's dashboard. Also assume that the mass of the body parts stopped by the knee is about 20% of the total body mass, which is 62 kg.

a) What is the minimum stopping time interval in needed to avoid injury to the knee if the person is initially traveling at 15 m/s (34 mi/h)? b) what is the minimum stopping distance?

Answer:

a) t= 46.5 msec. b) 0.35 m

Explanation:

Applying Newton´s 2nd Law to the mass supported by the knee (20% of the total mass), we can get the maximum acceleration allowable in order to avoid an injury, as follows:

a = F/m = 4000 N / 0.2*62 kg = 322.6 m/s²  

Applying the definition of acceleration, and taking into account that the knee finally come to an stop, we have:

a = vf - v₀ / Δt = -15 m/s / Δt

Solving for Δt :

Δt = -15 m/s / -322.6 m/s² = 0.0465 sec = 46.5 msec.

b) Assuming the acceleration remains constant during this time interval, we can find the distance needed to come to an ⇒stop, applying any of the kinematic equations, as this one:

vf² - v₀² = 2*a*Δx

⇒Δx = (vf²-v₀²) / 2*a

⇒Δx = -(15 m/s)² / 2*(-322.6 m/s²) = 0.35 m

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Oksana_A [137]

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Let's calculate force as we already have area;

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6 0
2 years ago
HELP WILL GIVE BRAINLIEST IF CORRECT
PolarNik [594]

Answer:

the answer is 1.35. Have a nice day!!

3 0
3 years ago
Read 2 more answers
AYUDAAA PORFAVOR
Sholpan [36]

Queremos crear un diagrama general para calcular el área de un triangulo.

Este será algo como:

  1. Definir variables
  2. Pedirle al usuario que introduzca los valores deseados (de las variables).
  3. Leer los valores deseados y asignarlo a la variable correspondiente.
  4. Realizar la operación para calcular el área.
  5. Mostrar en pantalla el resultado.

Como naturalmente habra algunas variaciones segun el programa que utilicemos, lo voy a escribir de forma bastante general.

Primero definamos nuestras variables:

Por ejemple, en fortran usariamos algo como:

real:: B, H, A

Donde B será la variable que usaremos para la base, H para la altura, y A para el área.

Luego tenemos que escribir en pantalla algo que le diga al usario que debe introducir la base y el area.

Luego el programa debe ser capaz de leer ese input.

con algo de la forma:

B = read*input 1

H = read*input 2

Una vez tenemos definidas las variables, simplemente calculamos el área del triangulo:

A = H*B/2

Finalmente la podemos mostrar en pantalla con algo como:

print(A).

Lo que nos mostraría el valor del área.

Concluyendo, el diagrama en general sería:

  1. Definir variables
  2. Pedirle al usuario que introduzca los valores deseados (de las variables).
  3. Leer los valores deseados y asignarlo a la variable correspondiente.
  4. Realizar la operación para calcular el área.
  5. Mostrar en pantalla el resultado.

Si quieres aprender más, puedes leer:

brainly.com/question/21949109

5 0
2 years ago
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