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Marina CMI [18]
3 years ago
13

(a+b)

rmula">
Physics
1 answer:
yulyashka [42]3 years ago
5 0
U distribute it
(a+b) • x2
ax2 + bx2
THE ANSWER IS ax2 + bx2 (the two means squared)
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It is 4.0 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h
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Answer:

Explanation:

To solve this problem we have to take into account that the energy consumed per second is the power. Hence, by multipling the power and the time spent to arrive to the lab we obtain the total energy consumed.

But first we have to calculate the time

t_{1}=\frac{x}{v_{1}}=\frac{4km}{10\frac{km}{h}}=0.4h=0.4(3600s)=1440s\\t_{2}=\frac{x}{v_{2}}=\frac{4km}{3\frac{km}{h}}=1.3h=1.3(3600s)=4800s\\

Now we use E=W*t for both times

E_{1}=t_{1}W_{1}=(1440s)(700W)=1008000J\\E_{2}=t_{2}W_{2}=(4800s)(290W)=1392000J\\

A. Hence, by running the energy consumed is lower.

B.

E1=1008000J

E2=1392000J

C. Because the more intense exercise is made in a lower time in comparison with the less intense exercise, and higher the time, more energy is consumed.

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3 years ago
What are the factors that affect ocean density?<br><br> Please help!!
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The density of seawater plays a vital role in causing ocean currents and circulating heat because of the fact that dense water sinks below less dense. long story short, seawater is the problem because its denser than pure water.

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A coil has an inductance of 6.00 mH, and the current in it changes from 0.200 A to 1.50 A in a time interval of 0.300 s. Find th
Andre45 [30]

Answer:

0.026 V

Explanation:

Given that,

Inductance of the coil, L = 6 mH

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We need to find the magnitude of the average induced emf in the coil during this time interval. The formula for the induced emf is given by :

\epsilon=L\dfrac{dI}{dt}\\\\=6\times 10^{-3}\times \dfrac{1.5-0.2}{0.3}\\\\=0.026\ V

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This energy would have converted to Kinetic. Write out an equation and the masses will cancel out. Does that hint help you to find the solution? If not, I will give you another hint.

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A 70 watt light bulb is run by a 120 v circuit. how many amps does it use
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