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Daniel [21]
3 years ago
6

It is 5.0 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h

(which uses up energy at the rate of 700 W), or you could walk it leisurely at 3.0 km/h (which uses energy at 290 W). Which choice would burn up more energy, and how much energy (in joules) would it burn? Why does the more intense exercise burn up less energy than the less intense exercise?
Physics
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer:

Walking burns up more energy,1740000J

Explanation:

Given that the displacement is 5.0km, and running at 10km/h and uses, walking at 3km/hr and uses 290watts:

Energy consumption for running is calculated as:

700watts=700j/s,d=5000m,v=\frac{25}{9}m/s\\\\\therefore E_c=\frac{d}{v}\times Energy \ usage\\\\=\frac{5000}{\frac{25}{9}}\times 700j/s\\\\=1260000J

Energy consumption for walking is calculated as:

290watts=290j/s,d=5000m,v=\frac{5}{6}m/s\\\\\therefore E_c=\frac{d}{v}\times Energy \ usage\\\\=\frac{5000}{\frac{5}{6}}\times 290j/s\\\\=1740000J

Walking is a slower process hence the need for more energy over longer periods  raltive to running the same distance.

Hence walking burns more energy; 1,740,000J. It burns more because you walk for a greater period of time.

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sertanlavr [38]
V=(40km/hr)(hr/3600s)(1000000mm/km)
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in a car moving at constant acceleration, you travel 230 mm between the instants at which the speedometer reads 40 km/hkm/h and
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The acceleration of the car is 0.8049m/s^{2}.It takes 13.802s to travel the 230 m.

<h3>What is acceleration?</h3>

In mechanics, acceleration refers to the rate at which an object's velocity with respect to time varies. Acceleration is a vector quantity (in that they have magnitude and direction). The direction of an object's acceleration is determined by the direction of the net force acting on it. Newton's Second Law states that the combined effect of two factors determines how much an item accelerates: 

(i) It follows that  the magnitude of the net balance of all external forces acting on the object is directly proportional to the magnitude of this net resulting force, and

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Calculations:

40 km/hr ----- 11.11m/s

80 km/hr ----- 22.22m/s

v^{2} -u^{2} =2as\\22.22^{2} - 11.11^{2} = 2 x a x 230\\ 370.296=460a\\ a= 0.8049m/s^{2} \\

Time taken

v-u=at

22.22-11.11= 0.8049 x t

t=13.802s

To learn more about acceleration ,visit:

brainly.com/question/2303856

#SPJ4

4 0
1 year ago
Solve the equation x=3logy2 for y.
melisa1 [442]

X = 3 · log(Y²)

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6 0
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GuDViN [60]

Answer: Technician B is right.

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3 0
3 years ago
A 0.73-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar
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Answer:

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B) to determine which side of the bar is positive, we must use the right hand rule

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Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West

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