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solniwko [45]
3 years ago
5

Waves travel as "packets" of several waves. in these "packets," a wave travels at __________.

Physics
2 answers:
dalvyx [7]3 years ago
8 0

Answer:

a. the same speed as the wave energy

Explanation:

Wave packet is only an envelope of a more localized wave. The entire wave is a combination of the discrete wave packets.

Wave is any disturbance that travels across space. A wave does not carry the particles of the medium in which travels. It carries energy from one point to another  as it travels.

For an electromagnetic wave, the energy of the wave is due to the vibrations of the electric and magnetic fields of the wave. For sound waves, wave energy is the vibrations or disturbance of the particles of the medium in which the sound wave travels.

madam [21]3 years ago
4 0
A. the same speed as the wave energy 
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This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e
Volgvan

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

P = \frac{1}{2} \rho v^3 A C_p

where \rho = 1.2 kg/m^3 is the air density, v = 8.89 m/s is the wind speed, A is the swept area and C_p = 0.45 is the efficiency

10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45

10000 = 190A

A = 10000 / 190 = 52.7 m^2

The swept area is a circle with radius r being the blade length

\pi r^2 = A = 52.7

r^2 = 52.7 / \pi = 16.79

r = \sqrt{16.79} = 4.1 m

4 0
3 years ago
The formula shown below is used to calculate the energy released when a specific quantity of fuel is burned. Calculate the energ
olya-2409 [2.1K]

Answer: 8400 J

Explanation:

The formula referenced in the question is:

Q=m. c. \Delta T  

Where:

Q  is the thermal energy

m=100g \frac{1 kg}{1000 g}=0.1 kg is the mass  of the water sample

c=4200 \frac{J}{kg\°C}  is the specific heat capacity of  water

\Delta T=20\°C  is the variation in temperature

Solving:

Q=(0.1 kg)(4200 \frac{J}{kg\°C})(20\°C)  

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Answer:

1000 Joules per second.

Explanation:

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W = mg\Delta h=120kg\cdot 10\frac{m}{s^2}\cdot 5m = 6000J

Power is work over the amount of time:

P = \frac{W}{\Delta t}=\frac{6000J}{6 s} = 1000 \frac{J}{s}

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