Una persona de 80kg intenta bajar de peso subiendo una montaña para quemar el equilante a una rebanada de pastel de chocolate (700kcal) ¿a que altura debe subir?

Explanation:

An 80kg person tries to lose weight by climbing a mountain to burn the equivalent of a large slice of a rich chocolate cake (700kcl). How high should you climb?

Given that,

A person of mass

M = 80kg

The person is climbing a stair case and he burn an energy of 700 kcl

We know that 1kcl = 4184J

Then, 700kcl = 700 × 4184J

700kcl = 2.9288 × 10^6 J

We want to find height of stairs the person will climb when he Loss 700kcl. The energy he burns will be converted to the potential energy

P.E is calculated using

P.E = mgh

Where

m is mass in kg

g is gravitational constant g = 9.81m/s2

h is the height of stairs he climb in metre

Then,

P.E = energy burn

mgh = 700kcl

80 × 9.81 × h = 2.9288 × 10^6

h = (2.9288 × 10^6) / (80 × 9.81)

h = 3731.91m

So, the person will climb 3731.91m before he losses 700kcl

In espanyol

Dado que,

Una persona de masa

M = 80 kg

La persona está subiendo una escalera y quema una energía de 700 kcl.

Sabemos que 1kcl = 4184J

Entonces, 700kcl = 700 × 4184J

700kcl = 2.9288 × 10 ^ 6 J

Queremos encontrar la altura de las escaleras que la persona subirá cuando pierda 700kcl. La energía que quema se convertirá en energía potencial.

P.E se calcula utilizando

P.E = mgh

Dónde

m es masa en kg

g es constante gravitacional g = 9.81m / s2

h es la altura de las escaleras que sube en metros

Entonces,

P.E = quema de energía

mgh = 700kcl

80 × 9.81 × h = 2.9288 × 10 ^ 6

h = (2.9288 × 10 ^ 6) / (80 × 9.81)

h = 3731.91m

Entonces, la persona subirá 3731.91m antes de perder 700kcl

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

the maxium height is 34m/s