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balandron [24]
3 years ago
9

What is the potential energy of a 2,000-kg car parked at the top of a 30-m hill?

Physics
2 answers:
babymother [125]3 years ago
5 0
Potential energy is a relative measure, so the answer is dependent on the assumptions we make. The potential energy in the car is going to be gravitational potential energy(PE). PE = mgh, where m is the mass, g is 9.8 m/s^2, and h is the height. So PE = 2000*9.8*h = 19600h. The final answer obviously depends on h. Most likely the problem is assuming that 30 meters under the top of the hill is considered 0 meters. Then h would be 30m and PE would equal 588 kJ.
vampirchik [111]3 years ago
5 0

Answer:

the answer is 588,000 J

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spin [16.1K]

To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,

R_{eq}= \frac{V}{I}

Here,

V = Voltage

I = Current

While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,

P=VI \rightarrow I = \frac{P}{V}

I =0.608 A

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R_{eq} = \frac{120V}{0.608A}

R_{eq} = 197.4\Omega

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3 years ago
An important aspect of fission reactions is that they produce _______________ which allow for chain reactions.
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An important aspect of fission reactions is that they produce free neutrons,  which causes chain reactions.

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3 years ago
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The gravitational force,F, on a rocket at a distance,r, from the center of the earth isgiven byF=kr2wherek= 1013N·km2. (Newton·k
Brrunno [24]

Answer:

The gravitational force changing velocity is

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Explanation:

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Differentiate the above equation

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The velocity is the distance in at time so

V=\frac{dr}{dt}=0.4 \frac{km}{s}

\frac{dF}{dt}=\frac{-2*k}{r^{3}}*0.4\\\frac{dF}{dt}=\frac{-8*10x^{13}N*km^{2} }{(10x10^{4}) ^{3}} \\\frac{dF}{dt}=\frac{-8x10^{12} }{1x10^{12}}

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8 0
3 years ago
A 0.7 kg block attached to a spring with force constant 160 Nm is free to move on a frictionless, horizontal surface. The block
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Answer:

+ 24 N

Explanation:

the computation is shown below:

Given that

Mass of the block = m = 0.7 kg

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x = 0.15m

Now the force on the block is

F = kx

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= 24 N

As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction

Therefore the third option is correct

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