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Goshia [24]
3 years ago
5

Martha must carry a 45 N package up three flights of stairs. Each flight of stairs has a height of 2.3m, and the actual distance

of a diagonal path she walks up the stairs is 10 m. What is the total work done on the package?
Physics
1 answer:
Darina [25.2K]3 years ago
3 0

Answer:

310.5 J

Explanation:

The total work done by Martha is equal to the increase in gravitational potential energy of the package, which is equal to

\Delta U = mg\Delta h

where

(mg) = 45 N is the weight of the package

\Delta h is the increase in height of the package

The package is carried up 3 flights of stairs, each one with a height of 2.3 m, so the total increase in heigth is

\Delta h = 3 \cdot 2.3 m=6.9 m

And so, the work done by Martha is

U=(45 N)(6.9 m)=310.5 J

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I am pretty sure it is B.

Explanation:

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A student evaluates a weight loss program by calculating the number of times he would need to climb a 14.0 m high flight of step
Liula [17]

Answer:

400 trips

Explanation:

Mechanical energy needed to climb 14 m by a man of 68 kg

= mgh

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= 9330 J

1 Kg of fat releases 3.77 x 10⁷ J of energy

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22% is converted into mechanical energy

so 22% of 1.6965 x 10⁷ J

= 3732.3 x 10³ J of mechanical energy will be available for mechanical work.

one trip of climbing of 14 m requires 9330 J of mechanical energy

no of such trip possible with given mechanical energy

= 3732.3 x 10³ / 9330

= 400 trips

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3 years ago
Como sacar velocidad final sin tener el tiempo
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A railroad car having a mass of 17.5 Mg is coasting at 1.5 m/s on a horizontal track. At the same time another car having a mass
Leokris [45]

Answer:

Then final velocity of the coupled system of cars will be 0.58 m/s

Explanation: Let us consider the car-1 is moving towards right (say + ve direction) and car-2 is moving towards left (say in - ve direction), accordingly velocity are considered +ve and -ve.

m_{1} = 17.5 \ Mg = 17.5 \times 10^{6} \ g = 17.5 \times 10^{3} \ Kg = 17500 \ Kg

v_{1} = + \ 1.5 \ m/s

m_{2} = 12 \ Mg = 12 \times 10^{6} \ g = 12 \times 10^{3} \ Kg = 12000 \ Kg

v_{2} = - \ 0.75 \ m/s

Applying the conservation of momentum, and let the final velocity of combined system is V m/s

m_{1} \times v_{1} + m_{2} \times v_{2} = (m_{1} + m_{2}) \times V

17500 \times 1.5 \  + 12000 \times (-0.75) = 29500 \times V

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V = \frac{17250}{29500} = 0.58 m/s

Then final velocity of the coupled system of cars will be 0.58 m/s

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