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FrozenT [24]
3 years ago
11

Why are light intensity, carbon dioxide concentration and temperature liimiting factors​

Physics
1 answer:
Harman [31]3 years ago
8 0

Answer:

As carbon dioxide concentrations increase, so too does the rate of photosynthesis until a certain point where the graph levels off. At lower carbon dioxide concentrations carbon dioxide is the limiting factor because an increase in carbon dioxide causes an increase in photosynthesis.

Explanation:

Hope it helps u

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What is the smallest possible piece an element in broken down into my normal laboratory processes
lozanna [386]

answer would be: A. atom

8 0
3 years ago
A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t
Mumz [18]

Answer:

v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t

Second stone:

y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

6 0
3 years ago
Read 2 more answers
Assume a solar cell would produce a current of 500mA and 3 V. What is the power of this solar cell?
defon

Answer:

the power of the solar cell is 1.5 watts

Explanation:

Recall that power is defined as the product of the voltage (V) times the running current (I): Power = V * I.

The only thing we have to take care of before actually performing the operation, is to convert milliamps into Amps, so our answer comes directly in the appropriate units (Watts). 500 mAmps can be written as 0.5 Amps, then, the product becomes:

Power = V * I = 3 V * 0.5 Amps = 1.5 watts

8 0
3 years ago
Explain Sound level intensity with mathematical steps?
yan [13]

Answer:

Sound intensity levels are quoted in decibels (dB) much more often than sound intensities in watts per meter squared. Decibels are the unit of choice in the scientific literature as well as in the popular media. The reasons for this choice of units are related to how we perceive sounds. How our ears perceive sound can be more accurately described by the logarithm of the intensity rather than directly to the intensity. The sound intensity level β in decibels of a sound having an intensity I in watts per meter squared is defined to be β(dB)=10log10(II0)β(dB)=10log10⁡(II0), where I0 = 10−12 W/m2 is a reference intensity. In particular, I0 is the lowest or threshold intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz. Sound intensity level is not the same as intensity. Because β is defined in terms of a ratio, it is a unitless quantity telling you the level of the sound relative to a fixed standard (10−12 W/m2, in this case). The units of decibels (dB) are used to indicate this ratio is multiplied by 10 in its definition. The bel, upon which the decibel is based, is named for Alexander Graham Bell, the inventor of the telephone.

Table 1. Sound Intensity Levels and IntensitiesSound intensity level β (dB)Intensity I(W/m2)Example/effect01 × 10–12Threshold of hearing at 1000 Hz101 × 10–11Rustle of leaves201 × 10–10Whisper at 1 m distance301 × 10–9Quiet home401 × 10–8Average home501 × 10–7Average office, soft music601 × 10–6Normal conversation701 × 10–5Noisy office, busy traffic801 × 10–4Loud radio, classroom lecture901 × 10–3Inside a heavy truck; damage from prolonged exposure[1]1001 × 10–2Noisy factory, siren at 30 m; damage from 8 h per day exposure1101 × 10–1Damage from 30 min per day exposure1201Loud rock concert, pneumatic chipper at 2 m; threshold of pain1401 × 102Jet airplane at 30 m; severe pain, damage in seconds1601 × 104Bursting of eardrums

8 0
3 years ago
what is the momentum of a child and a wagon if the total mass of the child and wagon is 22kg and the velocity is 1.5m/s?
antoniya [11.8K]

Answer:

33 kg m/s

Explanation:

The momentum of an object is given by:

p=mv

where

m is the mass of the object

v is the velocity of the object

In this problem, the total mass of the child and the wagon is m =22 kg, while the velocity is v = 1.5 m/s, therefore the momentum is

p=mv=(22 kg)(1.5 m/s)=33 kg m/s

7 0
3 years ago
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