Would the energy of the wave increase or decrease if the speed of the wave increases? Why?

1 answer:

4 0

Answer: The energy of a wave increases with an increase in the amplitude of the wave. The energy of a wave decreases with an increase in the amplitude of a wave.

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Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth

zalisa [80]

Refer to the diagram shown below.

Given:
R = 6.37 x 10⁶ m, the radius of the earth
h = 3.58 x 10⁷ m, the height of the satellite above the earth's surface.
Therefore
R + h = 4.217 x 10⁷ m

In geosynchronous orbit, the period of rotation is 1 day.
Therefore the period is
T = (24 h)*(60 min/h)*(60 s/min) = 86400 s

The angular velocity is
ω = (2π rad)/(86400 s) = 7.2722 x 10⁻⁵ rad/s

Part (a)
The tangential speed is
v = (R+h)*ω
= (4.217 x 10⁷ m)*(7.2722 x 10⁻⁵ rad/s)
= 3066.7 m/s
= 3.067 km/s

Part (b)
The centripetal acceleration is
a = v²/(R+h)
= (3066.7 m/s)²/(4.217 x 10⁷ m)
= 0.223 m/s²

Answers:
(a) The speed is 3.067 km/s
(b) The acceleration is 0.223 m/s²

7 0

3 years ago

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Roman55 [17]

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6 0

2 years ago

what would you want the after life to be like. examples are heaven and hell ,reincarnation ,eternal darkness , reliving your liv

NeTakaya

Answer:

i would want to be a dog or a cat

Explanation:

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4 0

2 years ago

A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00

gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as: $W=Q=Q_{in}-Q_{out}$ solving to Q_out we get: $Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ)$ this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: $n=1-\frac{T_{Low} }{T_{High}}$ where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get: $n=1-\frac{423.15}{723.15} =0.415=41.5%$