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Bingel [31]
3 years ago
13

How many grams of butter, which has a usable energy content

Physics
1 answer:
VARVARA [1.3K]3 years ago
4 0

Answer:

m_{butter} = 156.93 grams

Explanation:

Given data;

energy content  = 6.0 Cal/g = 6000 cal/g

mass of man 60 kg

climb height = 6.70 km

g = 9.80 m/s^2

we know that

potential energy = mgh

E_t = mU

wehre U - usable energy constant

equating both energy

60 \times 9.8 \times 6.70\times 10^3 = m_{butter} 6000\times \frac{4.184 j}{cal}

m_{butter} = 156.93 grams

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A car dropped from a height of 44 meters fall to a height of zero meters. How fast will the car be traveling as it hits the grou
FinnZ [79.3K]
Vi=0m/s
Vf=?
A=9.81
D=44
T=not needed

Vf^2=Vi^2+2ad
Vf=2ad square rooted
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Vf=29.3m/s
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vichka [17]
The answer to the problem b.
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3 years ago
A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci
Vlad [161]

Answer:

Approximately \rm 2.5\; m \cdot s^{-1}.

Explanation:

Let the increase in the rocket's velocity be \Delta v. Let v_0 represent the initial velocity of the rocket. Note that for this question, the exact value of  v_0 doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

  • Mass of the rocket with the 5 kg of fuel: 1000.
  • Initial velocity of the rocket and the fuel: v_0.
  • Hence the initial momentum of the rocket: 1000\,v_0.
  • Mass of the rocket without that 5 kg of fuel: 1000 - 5 = 995.
  • Final velocity of the rocket: v_0 + \Delta v.
  • Hence the final momentum of the rocket: 995\,(v_0 + \Delta v).
  • Mass of the 5 kg of fuel: 5.
  • Final velocity of the fuel: v_0 - 500 (assuming that the the 500 m/s in the question takes the rocket as its reference.)
  • Hence the final momentum of the fuel: 5\,(v_0 - 500).

Momentum is conserved in an isolated system like the rocket and its fuel. That is:

Sum of initial momentum = Sum of final momentum.

1000\,v_0 = 995\,(v_0 + \Delta v) + 5\,(v_0 - 500).

Note that 1000\, v_0 appears on both sides of the equation. These two terms could hence be eliminated.

0 = 995\, \Delta v - 5\times 500.

\displaystyle \Delta v = \frac{5}{995}\times 500 \approx \rm 2.5\; m \cdot s^{-1}.

Hence, the velocity of the rocket increased by around 2.5 m/s.

5 0
3 years ago
El conductor de un tren que circula a 20 m/s ve un obstáculo en la vía y frena con una aceleración de 2 m/s2 hasta parar ¿cuánto
AlladinOne [14]
Velocidad inicial = 20 m/s
velocidad final = 0 m/s
aceleracion = -2 m/s^2

aceleracion = (cambio de velocidad)/(cambio de tiempo)
(cambio de tiempo)= (cambio de velocidad)/aceleracion
tiempo = (-20 m/s)/(-2 m/s^2)
= 10 segundos

x = (x(inicial)) + (v(inicial))(tiempo) + 1/2(aceleracion)(tiempo)^2
x(inicial) = 0
x = (20 m/s)(10 s) + 1/2 (-2m/s^2)(10 s)^2
x = 200 m - 100 m
x = 100 m (el espacio recorrido en los dos segundos)

espero que esto te ayude! buena suerte!
6 0
3 years ago
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