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S_A_V [24]
3 years ago
14

Comparing energy resources

Physics
1 answer:
shtirl [24]3 years ago
4 0
You’re gonna wanna circle the first 5 at the top and do the bottom ones last
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The _____ is locked into a gravitational pull with our Milky Way, so both are actually
Lady bird [3.3K]

Answer:

Andromeda galaxy is the correct answer.

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2 years ago
A chemical reaction is when two elements or radicals change places with two other elements or radicals is a ... reaction
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That would be double replacement
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4 years ago
A pressure is applied to 2L of water. The volume is observed to decrease to 18.7 L. Calculate the applied pressure.
lakkis [162]

Answer:

Applied pressure is 1.08 10⁵ Pa

Explanation:

This exercise is a direct application of Boyle's law, which is the application of the state equation for the case of constant temperature.

PV = nR T

If T is constant, we write the expression for any two points

Po Vo = p1V1

From the statement the initial pressure is the atmospheric pressure 1.01 10⁵ Pa, so we clear and calculate

1 Pa = 1 N / m2

P1 = Po Vo / V1

P1 = 1.01 10⁵ 20/18.7

P1 = 1.08 10⁵ Pa

4 0
3 years ago
A concert loudspeaker suspended high off the ground emits 32.0 W of sound power. A small microphone with a 1.00 cm^2 area is 52.
lesya [120]

Answer:

Sound Intensity at microphone's position is 9.417\times 10^{- 4} W/m^{2}

The amount of energy impinging on the microphone is 9.417\times 10^{- 8} W/m^{2}

Solution:

As per the question:

Emitted Sound Power, P_{E} = 32.0 W

Area of the microphone, A_{m} = 1.00 cm^{2} = 1.00\times 10^{- 4} m^{2}

Distance of microphone from the speaker, d = 52.0 m

Now, the intensity of sound, I_{s} at a distance away from the souce of sound follows law of inverse square and is given as:

I_{s} = \frac{P_{E}}{Area} = \frac{P_{E}}{4\pi d^{2}}

I_{s} = \frac{32.0}{4\pi (52.0)^{2}} = 9.417\times 10^{- 4} W/m^{2}

Now, the amount of sound energy impinging on the microphone is calculated as:

If I_{s} be the Incident Energy/m^{2}/s

Then

The amount of energy incident per 1.00 cm^{2} = 1.00\times 10^{- 4} m^{2} is:

I_{s}(1.00\times 10^{- 4}) = 9.417\times 10^{- 4}\times 1.00\times 10^{- 4} = 9.417\times 10^{- 8} J

7 0
3 years ago
A person touches a large chunk of ice with their hand and remarks, “This is making me cold.” Explain what this person is feeling
malfutka [58]

Cold temperature can't "flow" into a warmer object, in this case, the person's hand. The heat energy of the hand gets transferred to the colder object, the ice.

7 0
4 years ago
Read 2 more answers
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