Investigators are working on a case where they need to know whether a watch will stop when it is dropped. In order to have a ver
2 answers:
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Answer:
Explanation:
<u>Given Data:</u>
Length = l = 820 mm = 0.82 m
Acceleration due to gravity = g = 9.8 ms⁻²
<u>Required:</u>
Frequency = f = ?
<u>Formula:</u>
<u>Solution:</u>
Answer:
The definition of acceleration is a change in the rate of motion, speed or action.
<h2>hope it helps.</h2><h2>stay safe healthy and happy..</h2>
Answer: 25.38 m/s
Explanation:
We have a straight line where the car travels a total distance , which is divided into two segments
:
(1)
Where
On the other hand, we know speed is defined as:
(2)
Where is the time, which can be isolated from (2):
(3)
Now, for the first segment the car has a speed
, using equation (3):
(4)
(5)
(6) This is the time it takes to travel the first segment
For the second segment the car has a speed
, hence:
(7)
(8)
(9) This is the time it takes to travel the secons segment
Having these values we can calculate the car's average speed :
(10)
(11)
Finally:
Answer:
a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.
b) Average power
P(w)= 1062.07 [w]
P(hp)=1.42 [hp]
c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.
Explanation:
First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)
Work:
Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )
Converting from Joules to Kcals:
Now lets take into account the efficiency of the human body (20%)
2.537 ---> 20%
x ---> 100%
So the student is consuming 12.685 KCals each time he runs up the stairs.
Now,
1 g --> 9 Kcals
1000 g --> 9000KCals
Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:
He must run up the stairs 709.5 times, to burn 1 Kg of fat.
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For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)
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