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3 years ago

Which "spheres" are interacting when water evaporates from plants

2 answers:

8 0

The plant grows in the solid part of earth, the lithosphere. When water evaporates from the plant, it enters the hydrosphere, the portion if earth on kand and in the air that contains water. The atmosphere is part of the hydrosphere.

Volgvan3 years ago

4 0

Answer:

Explanation:

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A wave with a frequency of 60 Hz is traveling along a string whose linear mass density is 230 g/m and whose tension is 65 N. If

matrenka [14]

To develop this problem we will use the concepts related to Speed in a string that is governed by Tension (T) and linear density (µ)

$V = \sqrt{\frac{T}{\mu}}$

Our values are given as:

$f = 60Hz\\\mu = 230 g/m = 0.230kg/m\\T = 65N\\P = 75w$

Replacing we have that the velocity is

$V = \sqrt{\frac{T}{\mu}}$

$V = \sqrt{\frac{65}{0.230}}$

$V = 16.81m/s$

From the theory of wave propagation the average power wave is given as

$P =\frac{1}{2} \mu \omega^2 A^2 V$

Where,

A = Amplitude

$\omega = 2\pi f \rightarrow$ Angular velocity

$A^2 = \frac{2P}{\mu \omega^2 V}$

$A^2 = \frac{2P}{\mu (2\pi f)^2 V}$

Replacing,

$A^2 = \sqrt{\frac{2(75)}{(0.230)(2\pi 60)^2(16.81)}}$

$A = 0.0165m$

Therefore the amplitude of the wave should be 0.0165m

8 0

3 years ago

If copper (E⁰ = ⁺0.34) and lithium (E⁰ = ⁻3.05) are used to make a battery, will electrons flow from lithium to copper or vice v

Lemur [1.5K]

Answer:

give me brainliest first and ill give u the correct anwers

Explanation:

4 0

2 years ago

3.5 g of a hydrocarbon fuel is burned in a vessel that contains 250. grams of water initially at 25.00 C. After the combustion,

kherson [118]

Answer:

3.) 463 J/g

Explanation:

Heat given by the fuel is used to increase the temperature of the water

so it is given as

$Q = ms\Delta T$

here we will have

m = 250 g

$\Delta T = 26.55 - 25$

now we also know that

$s = 4186$

so we have

$Q = (0.250)(4186)(26.55 - 25)$

$Q = 1622.075 J$

so 3.5 g fuel gives above energy

so per gram of fuel will have total energy given as

$Q = \frac{1622.075}{3.5}$

$Q = 463.45 J/g$

6 0

3 years ago

Sound level B in decibels is defined as

Lelechka [254]

Answer:

The approximate combined sound intensity is $I_{T}=1.1\times10^{-4}W/m^{2}$

Explanation:

The decibel scale intensity for busy traffic is 80 dB. so intensity will be

$10log(\frac{I_{1}}{I_{0}} )=80$ , therefore $I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}$

In the same way for the loud conversation having a decibel intensity of 70 dB.

$10log(\frac{I_{2}}{I_{0}} )=70$ , therefore $I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}$

Finally we add both of them $I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}$ , is the approximate combined sound intensity.

3 0

3 years ago

umka2103 [35]

You can hear a difference between these two sounds. That is because their pitch isdifferent. Pitch depends on the frequency of a sound wave. ... High sounds have highfrequencies and low sounds have lowfrequencies.

4 0

4 years ago