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Furkat [3]
2 years ago
7

Power lifter Paul lifts a 700.0kg barbell 2.00m in 0.400s. How much power did Paul develop?

Physics
2 answers:
Vsevolod [243]2 years ago
6 0

         Work = (force) x (distance.

The force required to lift the load is its weight.

         Weight = (mass) x (gravity)

so    Work = (mass) x (gravity) x (distance)

Now  Power = (work) / (time)

so     Power  =  (mass) x (gravity) x (distance) / (time)

                      =  (700kg) x (9.8 m/s²) x (2 m)  /  (0.4 sec)

                      =  ( 700 x 9.8 x 2) / (0.4)  (kg-m²/sec²) / (sec)

                      =        ( 34,300 )                  (joule)  /  (sec)

                      =                     34,300 watts .

This is one of those exercises where the math and the physics
are air-tight and bullet-proof but the answer is absurd.

34,300 watts is about 46 horsepower.  I don't care how many
Wheaties Power Lifter Paul had for breakfast today, he is NOT
snatching a barbell that weighs  1,543 pounds (0.77 ton !)
to the height of the top of his head in less than 1/2 second !
yawa3891 [41]2 years ago
3 0
34,335 W 
P= mgh/t 
P=700*2*9.81/.4

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Milk containing 3.7% fat and 12.8% total solids is to be evaporated to produce a product containing 7.9% fat. What is the yield
ratelena [41]

Answer:

the yield of product is YP=46.835 % and the concentration of solids is

Cs = 27.33%

Explanation:

Assuming that all the solids and fats remains in the milk after the evaporation, then the mass of product mP will be

Mass of fat in 100 kg of milk = 100 kg* 0.037 = mP* 0.079

mP = 100 kg* 0.037/0.079  =  46.835 kg

then the yield YP of the product is

YP= mP / 100 kg =  46.835 kg / 100 kg = 46.835 %

YP= 46.835 %

the concentration of solids Cs is

Mass of solids in 100 kg of milk = 100 kg* 0.128 = 46.835 kg * Cs

Cs = 100 kg* 0.128 / 46.835 kg  = 0.2733 = 27.33%

Cs = 27.33%

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6 0
2 years ago
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From a lake, water is pumped at a rate of 67 L/s to a storage tank positioned 14 m above while consuming 16.4 kW of electrical p
LenaWriter [7]

Answer:

57 %

Explanation:

input power = 16.4 kW = 16.4 x 10^3 W = 16400 W

Water pumped per second = 67 L/s

Mass of water pumped per second, m = Volume of water pumped epr second x density of water

m = 67 x 10^-3 x 1000 = 67 kg/s

height raised, h = 14 m

Output Power = m x g x h / t = 67 x 10 x 14 = 9380 W

efficiency = output power / input power = 9380 / 16400 = 0.57

% efficiency = 57 %

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3 0
3 years ago
car was moving in a straight road of length 320 km it covered 240 km with an average velocity 75 km/hr then it ran out of fuel a
Stella [2.4K]

The average velocity of the car for the whole journey is 69.57 km/h.

The given parameters:

  • <em>Length of the road, L = 320 km</em>
  • <em>Distance covered = 240 km at 75 km/h</em>
  • <em>time spent refueling, t₂ = 0.6 hr</em>
  • <em>Final velocity, = 100 km/hr</em>

The time spent by the before refueling is calculated as follows;

t = \frac{d}{v} \\\\t_1 = \frac{240}{75} \\\\t_1 = 3.2 \ hours

The time spent by the car for the remaining journey;

t_3 = \frac{320 - 240}{100} \\\\t_3 = 0.8 \ hr

The total time of the journey is calculated as follows;

t = t_1 + t_2 + t_3\\\\t = 3.2 \ hr \ + \ 0.6 \ hr \ + \ 0.8 \ hr\\\\t = 4.6 \ hours

The average velocity of the car for the whole journey is calculated as follows;

v = \frac{total \ distance }{total \ time} \\\\v = \frac{320}{4.6} \\\\v = 69.57 \ km/h

Learn more about average velocity here: brainly.com/question/6504879

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