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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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Work done=F×d

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Answers:

a) -171.402 m/s  

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final height

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb's initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity </h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

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3 years ago
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