2 years ago

Which title goes with this list: Amendments Articles Preamble?

Physics

1 answer:

anastassius [24]2 years ago

8 0

D. City and U.S. Government, but I’m not sure what the context is though.

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Belly-flop Bernie dives from atop a tall flagpole into a swimming pool below. His potential energy at the top is 7000 J (relativ

elena55 [62]

Answer:

KE₂ = 6000 J

Explanation:

Given that

Potential energy at top U₁= 7000 J

Potential energy at bottom U₂= 1000 J

The kinetic energy at top ,KE₁= 0 J

Lets take kinetic energy at bottom level = KE₂

Now from energy conservation

U₁+ KE₁= U₂+ KE₂

Now by putting the values

U₁+ KE₁= U₂+ KE₂

7000+ 0 = 1000+ KE₂

KE₂ = 7000 - 1000 J

KE₂ = 6000 J

Therefore the kinetic energy at bottom is 6000 J.

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2 years ago

A block with a mass of 9.00 kg is pulled at a constant speed across a horizontal tabletop with a spring scale. The scale reads 6

snow_tiger [21]

Answer:0.69

Explanation:

Coefficient of kinetic friction=f/R=61.8/90=0.69

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2 years ago

The compass is a small bar magnet that’s free to spin. Based on what you observed, is the coil of wire with current passing thro

OverLord2011 [107]

Answer:

yes, it is a magnet.

Explanation:

The electrons that are travelling through the wire go in 1 direction allowing it to create a magnetic field around its area.

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1 year ago

In a laser, electrical energy is changed to light energy, and a small amount of the energy is wasted as heat. Describe the total

tangare [24]

Bvcxczxcghchkgckghcgc energy to light

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2 years ago

The sound from a loud speaker has an intensity level of 80 db at a distance of 2.0m. Consider the speaker to be a point source,

Tamiku [17]

Answer:

2.83m

Explanation:

The information that we have is

Intensity at 2.0 m: $I=80dB$ and $r_{1}=2m$

we need an intensity level of: $I_{2}=40dB$

thus, we are looking for the distance $r_{2}$ .

which we can find with the law for intensity and distance:

$(\frac{r_{2}}{r_{1}} )^2=\frac{I_{1}}{I_{2}}$

we solve for $r_{2}$ :

$\frac{r_{2}}{r_{1}}=\sqrt{\frac{I_{1}}{I_{2}} }\\\\r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}} }$

and we substitute the known values:

$r_{2}=(2m)\sqrt{\frac{80dB}{40dB} }\\\\r_{2}=(2m)\sqrt{2}\\ r_{2}=2.83m$

at a distance of 2.83m the intensity level is 40dB

5 0

2 years ago