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stepan [7]
3 years ago
9

What is potential and kinetic energy

Physics
1 answer:
Galina-37 [17]3 years ago
3 0
<h2><u>Explanation:</u></h2>

______________________________

<u />

<h3>Potential:</h3>
  • This refers to an object having energy running through it, but being stored. It hasn't quite been released yet.
  • Ex: <em>Someone waiting to dive into a pool, when a whistle blows.</em>
<h3 /><h3>Kinetic: </h3>
  • This refers to the energy going through an object but that energy is moving, it's been released.
  • Ex: <em>A boy pedaling his bike atop a hill. </em>

______________________________

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2) A ray of light in air is approaching the boundary with water at an
KiRa [710]

Answer:

Explanation:

ASSUMING the 52° is the angle of incidence measured from the perpendicular to the surface

          n₁sinθ₁ = n₂sinθ₂

          1 sin52 = 1.33sinθ₂

                  θ₂ = arcsin(sin52 / 1.33)

                  θ₂ = 36°

as measured from the perpendicular to the surface

5 0
3 years ago
A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s
strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v

Where:

m_{M}, m_{S} - Masses of the small meteorite and the communication satellite, measured in kilograms.

v_{M}, v_{S} - Speeds of the small meteorite and the communication satellite, measured in meters per second.

v - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}

If m_{M} = 1\times 10^{-3}\,kg, m_{S} = 200\,kg, v_{M} = 20000\,\frac{m}{s} and v_{S} = 0\,\frac{m}{s}, the final speed is now calculated:

v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}

v = 0.1\,\frac{m}{s}

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

K_{S} + K_{M} - K - Q_{disp} = 0

Q_{disp} = K_{S}+K_{M}-K

Where:

K_{S}, K_{M} - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

K - Kinetic energy of the satellite-meteorite system, measured in joules.

Q_{disp} - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]

Given that m_{M} = 1\times 10^{-3}\,kg, m_{S} = 200\,kg, v_{M} = 20000\,\frac{m}{s}, v_{S} = 0\,\frac{m}{s} and v = 0.1\,\frac{m}{s}, the dissipated heat is:

Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]Q_{disp} = 200000\,J

Q_{disp} = 200\,kJ

The energy coverted to heat is 200 kilojoules.

4 0
3 years ago
A 20 kg mass is dropped from a tall rooftop and accelerates at 9.8 m/s2. What is the weight of the dropped object?
patriot [66]
Objects in free fall are weightless.
6 0
3 years ago
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A ___ is when too much water moves into an area
Marina86 [1]
A. Flood.
Hope it helps!
6 0
3 years ago
Read 2 more answers
Learning Goal: To understand the nature of a sound wave, including its properties: frequency, wavelength, loudness, pitch, and t
Lady_Fox [76]

Answer:

Sound wave is a longitudinal wave that propagates in a medium

Explanation:

<em>Part A:</em> (C) Sound wave is propagation of pressure fluctuations in a medium.

<em>Part B: </em>(C) Pressure fluctuations travel along the direction of propagation of         the sound wave.

<em>Part C: </em>(A) Yes air play a role in the propagation of the human voice from one end of the lecture hall to the other.

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3 years ago
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