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Mandarinka [93]
2 years ago
13

1. Why is only a small portion of the solar energy that strikes Earth's atmosphere stored by primary producers?

Physics
1 answer:
DanielleElmas [232]2 years ago
5 0

Explanation:

only a fraction of radiation strikes plants or algae and only a portion of that fraction is of wavelengths suitable for photosynthesis and much energy is lost as a result of reflection or heating of plant tissue

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Which diagram is the best representation of gas molecules in a closed container? A. Diagram A B. Diagram B C. Diagram C D. Diagr
WITCHER [35]
Gas "floats" so if there are examples or pictures it would be the one with the most evenly spread out "dots". 
8 0
3 years ago
If a car travels 30 mi. north for 30 min., 60 mi. east for 1.0 hour, and 30 mi. south for 30 min., what is the average speed
Ainat [17]

Answer: 60mph

Explanation:

Given the following :

First leg travel:

Distance = 30 miles

Time of travel= 30 minutes = 0.5 hour

Second leg travel:

Distance = 60 miles

Time of travel = one hour

Average speed :

Speed = total Distance / time of travel

Total distance in miles = (30 + 60) miles = 90 miles

Total time of travel = 1 hour + 0.5 hour = 1.5 hours

Average speed = total distance traveled / total travel time

Average speed = 90 miles / 1.5 hours

Average speed = 60 miles / hour

= 60mph

3 0
4 years ago
At which point or points does the pendulum have the greatest kinetic energy?
Daniel [21]

At the lowest point of its motion, kinetic energy is maximum and potential energy is minimum. This is where the velocity is a maximum. At the highest point of its motion, kinetic energy is minimum (i.e. zero) and potential energy is maximum.

4 0
3 years ago
Read 2 more answers
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
Two loudspeakers are placed next to each other and driven by the same source at 500 Hz. A listener is positioned in front of the
stellarik [79]

To solve this problem we will apply the concepts related to wavelength as the rate of change of the speed of the wave over the frequency. Mathematically this is

\lambda = \frac{v}{f}

Here,

v = Wave velocity

f = Frequency,

Replacing with our values we have that,

\lambda = \frac{340}{500}

\lambda = 0.68m

The distance to move one speaker is half this

\lambda/2 = 0.34m

Therefore the minimum distance will be 0.34m

7 0
3 years ago
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