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2 years ago

A weightlifter raises a 180-kg barbell to a height of 1.95 m. How much work is done by the weightlifter in lifting the barbells

Physics

1 answer:

melamori03 [73]2 years ago

8 0

P= 10m = 10 x 180 = 1800 N

work is done by the weightlifter in lifting the barbells :

A= Ph = 1800 x 1.95 = 3510 J

ok done. Thank to me :>

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A student is pedaling a stationary bicycle at the U of M Rec Center. Her alveolar PO2 = 115 mm Hg and her oxygen consumption is

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Answer:

This is an alveolar gas equation question and it is used to approximate the partial pressure of oxygen in the alveolus (PAO2):

The equation states;

PₐO₂ = P₁O₂ - (PₐCO₂/R) = [(PB - PH₂O) × F₁O₂ - (PₐCO₂/R)] ...................Eqn 1

where

PₐO₂ = Alveolar partial pressure of O2 = 115mmHg

P₁O₂ = Inspired partial pressure of O2 = 150mmHg

PB = barometric pressure,

PH₂O = Water vapor pressure (usually 747 mmHg),

F₁O₂ = fractional concentration of inspired oxygen,

and R = gas exchange ratio. (Usually around 0.8)

FₐO₂ = Fraction of alveolar O₂

(O₂) = 1L/min = 1dm³

From eqn 1. we have

PₐCO₂ = (P₁O₂ - PₐO₂)/R

= (150 - 115)x0.8

PₐCO₂ = 28mmHg

Similarly from Eqn 1, we have

F₁O₂ = (PₐO₂ + PₐCO₂/R)/(PB - PH₂O)

F₁O₂ = (115 + (28/.8))/(747 - 47)

F₁O₂ = 0.21

Now to find the Alveolar Ventilation A, we will use this equation;

O₂ = A(F₁0₂ - FₐO₂) .................Eqn 2

But FₐO₂ = PₐO₂/(PₐO₂ + PₐCO₂)

FₐO₂ = 115/ (115+28) = 0.8

A = O₂/(F₁0₂ - FₐO₂)

A = 0.001/(0.21 - 0.8)

A = 0.00169m³/min

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2 years ago

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Answer:it's a chemical change.physical changes occur without 2 or more reactants,a reaction,and a product

Explanation:

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3 years ago

A weightlifter uses a force of 356 N to lift a set of weights 2.2 m off the ground. How much work did the weightlifter do? Answe

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Answer:

783.2 N.m or Joules

Explanation:

$W=F*d*cos(Ф)$

Work done when a force F is applied to move an object by a displacement d meters

where cos(Ф) is the angle between applied force F and displacement d

Since the weightlifter is applying the force upward and the set of weight also move upward therefore, both are in same direction hence angle will be zero.

Now lets substitute the given values into the work equation:

$W=356*2.2*cos(0)$

since cos(0)=1

$W=356*2.2*1$

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$W=783.2 Joules$

Since the unit of Force is Newton (N) and the unit of displacement is meters (m) therefore, unit of work done will be N.m

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The value of the constant needed to complete in joules to three significant figures is 2.18 x 10^-18 J.

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For the first part, we are looking for Vf when dy=11.0
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For the part b, t is equals to the time took to reach and dy is equals to 11.0
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