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Margarita [4]
2 years ago
11

It is 6.00 km from your home to the physics lab. as part of your physical fitness program, you could run that distance at 10.0 k

m/hr (which uses up energy at the rate of 700 w ), or you could walk it leisurely at 3.00 km/hr (which uses energy at 290 w ).
Physics
1 answer:
kirill115 [55]2 years ago
7 0
<span>1.) It is 6.00km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10.0km/hr (which uses up energy at the rate of 700W ), or you could walk it leisurely at 3.00km/hr (which uses energy at 290 W). A.)Which choice would burn up more energy? running or walking? b.)How much energy (in joules) would it burn? c.)Why is it that the more intense exercise actually burns up less energy than the less intense one? Follow 2 answers Report Abuse Answers billrussell42 Best Answer: running, at 10 km/hour for 6 km is 6 km / 10 km/hour = 0.6 hour or 36 min energy used is 700 watts or 700 joules/s x 36 min x 60s/min = 1.512e6 joules or 1.5 MJ walking, at 3 km/hour for 6 km 6 km / 3 km/hour = 2 hour or 120 min energy used is 290 watts or 290 joules/s x 120 min x 60s/min = 1.872e6 joules or 1.8 MJ C) should be obvious PS, this has nothing to do with potential energy. billrussell42 · 5 years ago 0 Thumbs up 1 Thumbs down Report Abuse Comment Simon van Dijk I assume the watt consumption is per hour. Then running 6km at 10.0 km/h results in 700*6/10 = 420 w.h and walking in 290*6/3 = 580 w.h So walking would burn up more energy (kwh) b) 1 kilowatt hour = 3 600 000 joules so 420 wh = 0.42 kwh = 1.51.10^6 joule c) when you put more effort in making the distance your energy is used more efficient. Simon van Dijk · 5 years ago 0 Thumbs up 2 Thumbs down Report Abuse Comment</span>
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Write down two advantages of parallel combination ​
Simora [160]

Answer:

In parallel combination each appliance gets the full voltage.

If one appliance is switched on/of others are not affected.

The parallel circuit divide the current through the appliances.

In a parallel combination it is very easy to connect or disconnect a new appliance without affecting the working of other appliances.

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Explanation:

8 0
2 years ago
You observe a distant galaxy. You find that a spectral line, resulting from an electron transition in hydrogen, is shifted from
alexandr1967 [171]

Answer:

The galaxy is moving away from the observer

Explanation: when a galaxy is moving away from us, the light we percieve from it is "streched". Since the wavelength has an inverse raltionship whith frequency, the longer the wavelength is, the lower the frequency. And lower frequencies correspond to red and infrarred light.

So when we see the light has shifted to the infrarred part of the spectrum, it means the source is traveling away from us, making the light waves we percieve streched and move from visible light to infrarred.

6 0
2 years ago
If a diffraction grating has 3700 lines per cm, what is the spacing d between lines
Sonja [21]
So first we find the gap between the slits by the formula d=1/N 

<span>N is number of lines per metre so 3700 line/cm = 370000 lines/m </span>
<span>So d=2.7*10^-6 </span>

<span>Now we use the formula dsin(angle)=n(wavelength) </span>

<span>d is the same </span>
<span>n is the order of the diffraction pattern </span>

<span>so wavelenth=dsin(angle)/n </span>
<span>=[(2.7*10^-6)*sin30]/3 </span>
<span>=4.5*10^-7 m</span>
7 0
2 years ago
A 5.93 kg ball is attached to the top of a vertical pole with a 2.35 m length of massless string. The ball is struck, causing it
Delvig [45]

Answer

given,

mass of ball = 5.93 kg

length of the string = 2.35 m

revolve with velocity of 4.75 m/s

acceleration due to gravity = 9.81 m/s²

T cos θ = mg

T cos θ = 5.93\times 9.81

T cos θ = 58.17

T sin \theta =\dfrac{mv^2}{r}

T sin \theta =\dfrac{5.93\times 4.75^2}{2.35 sin \theta}

T sin^2 \theta =56.93

sin^2 \theta = 1 - cos^2 \theta

T (1 - cos^2 \theta) =56.93

T (1 - (\dfrac{58.17}{T})^2) =56.93

T² - 56.93T - 3383.75 = 0

T =  93.22 N

cos \theta = \dfrac{58.17}{93.22}

θ = 51.39°

6 0
3 years ago
Which describes the average velocity of an ant traveling at a constant speed
yanalaym [24]

Answer: A. The total displacement divided by the time and  C. The slope of the ant's displacement vs. time graph.

Explanation:

Hi! The question seems incomplete, but I found the options on the internt:

A. The total displacement divided by the time.

B. The slope of the ant's acceleration vs. time graph.

C. The slope of the ant's displacement vs. time graph.

D. The average acceleration divided by the time.

Now, since we know the ant is travelling at a constant speed, its average velocity V will be expressed by the following equation:

V=\frac{d}{t}

Where:

d is the ant's total displacement

t is the time it took to the ant to travel to the kitchen

Hence one of the correct options is: A. The total displacement divided by the time

On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is: C. The slope of the ant's displacement vs. time graph.

3 0
3 years ago
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