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Aloiza [94]
2 years ago
13

Using Newton's third law of motion, write a scientific explanation that describes why

Physics
1 answer:
Aneli [31]2 years ago
4 0

Answer:

when a force is applied by one object to a second object, an equal and opposite force is applied back on the first object

Explanation:

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A certain shade of blue has a frequency of 7.15 × 1014 hz. what is the energy of exactly one photon of this light?
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The energy carried by a single photon of frequency f is given by:
E=hf
where h=6.6 \cdot 10^{-34} m^2 kg s^{-1} is the Planck constant. In our problem, the frequency of the photon is f=7.15 \cdot 10^{14}Hz, and by using these numbers we can find the energy of the photon:
E=(6.6\cdot 10^{-34}m^2 kg s^{-1})(7.15 \cdot 10^{14}Hz)=4.7 \cdot 10^{-19}J
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3 years ago
The diagram shows the Earth rotating on it's axis. The two stars show different locations on the surface... How long does it tak
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My guess would be about 10 years because stars are hot balls of light that are reflections from years ago so it would most likely take awhile
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3 years ago
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The 10-kg uniform rod is pinned at end
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Supposing that the spring is un stretched when θ = 0, and has a toughness of k = 60 N/m.It seems that the spring has a roller support on the left end. This would make the spring force direction always to the left 
Sum moments about the pivot to zero. 
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3 years ago
Imagine that you're standing in a large room when a loud noise is made.You begin hearing a series of echoes.Which charactestics
e-lub [12.9K]
The correct is  Reverberation. A reverberation is created when a sound or signal is reflected causing a large number of reflections to build up and then decay as the sound is absorbed by the surfaces of objects in the space – which could include furniture, people, and air. 
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3 years ago
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A heat pump with a COP of 3.15 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the te
Jet001 [13]

Answer: 1026s, 17.1m

Explanation:

Given

COP of heat pump = 3.15

Mass of air, m = 1500kg

Initial temperature, T1 = 7°C

Final temperature, T2 = 22°C

Power of the heat pump, W = 5kW

The amount of heat needed to increase temperature in the house,

Q = mcΔT

Q = 1500 * 0.718 * (22 - 7)

Q = 1077 * 15

Q = 16155

Rate at which heat is supplied to the house is

Q' = COP * W

Q' = 3.15 * 5

Q' = 15.75

Time required to raise the temperature is

Δt = Q/Q'

Δt = 16155 / 15.75

Δt = 1025.7 s

Δt ~ 1026 s

Δt ~ 17.1 min

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