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2 years ago

Describe a scenario when you could use energy but not do any work.

Physics

1 answer:

Harlamova29_29 [7]2 years ago

4 0

Answer:

When you don't move, you still use energy. This energy is called potential energy, or, stored energy.

When you don't move or do work, you can use energy.

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What was the name of charles lindbergh’s single winged airplane for his solo flight across the atlantic?

aev [14]

Answer:

Spirit of St. Louis

Explanation:

Charles Lindbergh was known as a prolific aviator during the early twentieth century. He is well known for the flight he took from Long Island, New York, to Paris, France. It was a continuous flight across the Atlantic Ocean.

The plane he used was the Spirit of St. Louis which took more than 33 hours to complete the journey. It was the first successful flight of this kind. The airplane flew from Long Island on May 20 and landed in Paris on May 21.

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3 years ago

What are the three examples that show relation between pressure and area in daily life.<br>

makvit [3.9K]

Answer: The area of the edge of a knife's blade is extremely small.

Syringes are used to take blood for blood tests.

When air is sucked out of a drinking straw, the air pressure inside if decreases and the atmospheric pressure outside forces the liquid to go inside the straw.

Explanation:

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2 years ago

How do l calculate e<br> nergy

aivan3 [116]

The formula for energy of motion is KE = .5 x m x v^2

Ke= Kinetic Energy in Joules

m = Mass in Kilograms

v = Velocity in Meters per Second

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2 years ago

What is the necessary conditions for the production of sound?

TEA [102]

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Answer: Something that's vibrating, and you also need medium for those vibrations to start in.

I hope this helped!

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2 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