A power lifts 500 lb 6 ft off the ground. how much work has he performed?___ft-l
ID: 1414775 • Letter: A
Question
A power lifts 500 lb 6 ft off the ground. how much work has he performed?___ft-lb work =____kg-m work. How many calories of energy did he use to produce this work?____cal. If he performs this feat 11 times in 1 min. What is his power output?_____HP =_____W. A physician has ordered a 0.35 mg/kg of body weight drug dosage for a 215-lb patient to control his blood sugar levels. How large is each dose?____Atomic weights: Na = 25 k = 39 Ca = 40 Cl = 35.5 Molecular weights: Glucose = 180 NaCl = 58.5 KCl = 74 CaCl_2 = 110 How many grams of glucose would you need to make 500 ml of an 8% solution?____What is the percent concentration of 6 g of NaCl dissolved in 1 L of solution?_____How many grams of KCl would you need to make 250 ml of a 0.5-M solution?____Explanation / Answer
1st two answers are right.
1 ft-lb = 1.355 Joules
therefore 3000 ft-lb = 3000 x 1.355 = 4065 J
1 cal = 4.18 Joules
and so Work done in calories will be: 4065/4.18 = 972.5 cal.
Total Energy used up in 11 reps of this weight lift will be : 11 x 4065 J = 44,715 J
if this is done in 1 minute then the Power Output will be:
44,715/60 = 745.25 W approximately 1 Horse Power.
3) Dosage = 0.35mg/kgof body weight
1lb = 0.45kg
so 215lb = 215 x 0.45 = 96.75 kg
therefore the total dose in one administration will be : 0.35 x 96.75 = 33.8625mg
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