Why are fossils more commonly found in sedimentary rocks and not igneous rocks?
1 answer:
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Answer:
a) 1. 1365 × 10⁻⁹N
b) 9.1862 × 10⁻⁴⁶N
c) 8.61 × 10⁻¹¹C/Kg
Explanation:
Ke = 8.99× 109 N.m2 / C2 , G = 6.67 × 10-11 N. m2 / kg2
a) F = Ke . q₁.q₂/r²
= (8.99× 109 N.m2 / C2 ) ×(1.60×10⁻¹⁹C)²/(4.50×10⁻¹⁰C)²
= 1. 1365 × 10⁻⁹N
b)
F = Gm₁m₂/r²
= (6.67× 10⁻¹¹)×(1.67×10⁻²⁷)²/(4.50×10⁻¹⁰)²
= 9.1862 × 10⁻⁴⁶N
The electric force is larger by 8.0497 ×10³⁷ times
c)
if Keq₁q₂/r² Gm₁m₂/r²,
with q₁=q₂ = q, and m₁ =m₂ = m
Then q/m = =
= 8.61 × 10⁻¹¹C/Kg
Answer:
The answers to the questions are;
a. The number of times the student run the flight of stairs to lose 1.00 kg of fat is 829.23 times.
b. The average power output, in watts and horsepower, as he runs up the stairs is 158.026 watts.
c. The act of climbing the stairs is not a practical way to lose weight has to lose 1 kg of fat, the student needs to workout for about 26.49 hrs or 1.104 days.
Explanation:
To solve the question, we write out the known variables as follows
1 g of fat = 9.00kcal
Number of steps the student climbs = 95 steps
Height of each step = 0.150 m
Time it takes for the student to reach the top of the stairs = 57.5 s.
Efficiency of human muscles = 20 %
Mass of student, m = 65 kg
a. From the question, the energy expended by the student in climbing the stairs is the "work done" by the student.
The "work done" is the height climbed resulting in the gaining of gravitational potential energy P. E..
That is work done, W, = P. E. = m·g·h
Where:
h = The total height climbed by the student
g = Acceleration due to gravity = 9.81 m/s²
Therefore;
h = Height of each step × Number of steps the student climbs =
= 0.150 m/(step) × 95 steps = 14.25 m
Therefore, P. E. = 65 kg × 9.81 m/s² × 14.25 m = 9086.5125 kg·m²/s²
= 9086.5125 J
We remember that the efficiency of the muscle is 20 %
The formula for efficiency is
Efficiency =
The work produced by the muscle = Energy Output = 9086.5125 J
Energy input is given by
= 9086.5125 J/ (0.2) = 45432.5625 J
= 45.432 kJ
From the question, 1 g of fat = 9.00 kcal and
1 kcal = 4186 J
Therefore 1 g of fat can release 9.00 kcal × 4186 J = 37674 J
Therefore 1 kg of fat = 1000 g = 1000 × 37674 J = 37674 kJ
To consume the energy in 1 kg of fat the student therefore will run up the foight of stairs times to make up the 37674 kJ energy contained in 1 kg of fat
That is = 829.23 times
b. Power is the rate of doing work
That is Power output = =
= 158.026 watts
c. No as the activity student will have to spend a total time of
829.23 × 57.5 s = 47680.67 s climbing up the stairs alone and
47680.67 s = 13.24 Hours climbing up of which if the time to climb down is the same s climbing up, then we ave total time = 2× 13.24 Hours
= 26.49 hrs = 1.104 days exercising which is not humanly possible.