The energy (in calories) required to power the heart for one day is 268.32 Cal
<h3>How to convert 1 day to seconds</h3>
1 day = 24 h
24 h = 24 × 60 = 1440 mins
1440 mins = 1440 × 60 = 86400 s
Thus,
1 day = 86400 s
<h3>How to determine the energy (in calories) for 1 day (86400 s)</h3>
From the question given,
1 s = 13 J
Therefore,
86400 s = 86400 × 13
86400 s = 1123200 J
Divide by 4186 to express in calories
86400 s = 1123200 / 4186
86400 s = 268.32 Cal
Thus, the energy needed to power the heart for 1 day is 268.32 Cal
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Complete question
Under normal conditions the human heart converts about 13.0 J of chemical energy per second into 1.30 W of mechanical power as it pumps blood throughout the body. (a) Determine the number of Calories required to power the heart for one day given that 1 Calorie equals 4186
Answer:
gravitational potenetial energy =ℎ
Explanation:
none
Answer:
both
Explanation:
really you don't know something from kindergarden
Answer:
the apparent weight of the astronaut is 81.032 N { towards moon }
Explanation:
Given that;
Mass of astronaut m = 80 kg
Distance of spaceship from the Earth's moon r = 2200 km = 2200 × 10³ m
Acceleration due to gravity of the moon = GM/r²
where M is mass of the moon( 7.34767309 × 10²² kg )
gravitational constant G = 6.67 × 10⁻¹¹
So,
Acceleration due to gravity of the moon g is;
g = [ (6.67 × 10⁻¹¹) × (7.35 × 10²²) ] / (2200 × 10³)²
g = 4.90245 × 10¹² / 4.84 × 10¹²
g = 1.0129 m/s²
now, we take the positive direction towards the moon if the spacecraft is moving with constant velocity, a = 0
The apparent weight is measured by the normal force FN
so,
∑F = ma
-FN + mg = ma
-FN + mg = 0
FN = mg
we substitute
FN = 80 × 1.0129
FN = 81.032 N { towards moon }
Therefore, the apparent weight of the astronaut is 81.032 N { towards moon }
Answer:
It depends upon what you have given in the question but according to Ohm's law,
where, V = voltage (v)
I = current (amperes)
R = resistance (Ω)
Other formulas are:
Hope it helps!<3