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Tema [17]
3 years ago
9

How much power is required to do 40 J of work on an object in 5 seconds?

Physics
1 answer:
m_a_m_a [10]3 years ago
6 0
C 
because 40 j dived by 5 seconds is 8 work
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A tall building is swaying back and forth on a gusty day. The wind picks up and doubles the amplitude of the oscillation. By wha
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by a factor of 2

Explanation:

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How do we convert grams to kilograms
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4 years ago
A biker goes out for a ride and begins riding North at 18 km/h for 200 minutes, then the driver turns around and rides at 12 km/
Juliette [100K]

Answer:

Explanation:

18 km / h

= 300 m / min

12 km / h = 200 m / min

distance travelled in 200 minutes = 300 x 200 = 60000 m

distance travelled in 50 minute in return journey = 200 x 50 = 10000 m

total distance travelled = 70000 m

total time = 250 minute

speed = 70000 / 250

= 280 m / min

= 16.8 km / h

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total time = 250 min

velocity = 50000 / 250

= 200 m / min

= 12 km / h

4 0
3 years ago
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
IgorC [24]

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 = \frac{Ene rgy Out put}{Energ y In put} \times 100 %

The work produced by the muscle =  Energy Output = 9086.5125 J

Energy input is given by

\frac{Out put} {Effici ency} = 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 \frac{37674 kJ}{45.432 kJ} times to make up the 37674 kJ energy contained in 1 kg of fat

That is  \frac{37674 kJ}{45.432 kJ} =  829.23 times

b. Power is the rate of doing work

That is Power output = \frac{ WorkO utput }{Time} = \frac{9086.5125 J}{57.5 s} = 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.

3 0
3 years ago
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