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mihalych1998 [28]
3 years ago
9

What kind of organisms are able to convert sunlight into usable energy?

Physics
2 answers:
Tamiku [17]3 years ago
4 0

Answer:

Photosynthetic organisms

Explanation:

The electromagnetic energy of sunlight is converted to chemical energy in the chlorophyll-containing cells of photosynthetic organisms. In eukaryotic cells these reactions occur in the organelle known as the chloroplast

Hope this helps! :)

MatroZZZ [7]3 years ago
3 0

Answer:

The producers

Explanation:

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Ionic bonds are formed between a cation (metal) and an anion (nonmetal)
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If the potential difference across a 12-ohm resistor is 6 volts, the current through the resistor is?
larisa86 [58]
Current is inversely proportional to the resistance of the resistor and directly to the potential difference across it.

I = V/R = 6/12 = 0.5 A
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3 years ago
Identify which best describes the energy used to pluck guitar strings to make sound.
Novay_Z [31]
<span>The correct answer is: Mechanical Energy

Explanation:
As the guitar strings are plunked, the potential energy stored in the strings has an ability to make them vibrate. When the strings are vibrating, that potential energy is actually converted to the kinetic energy. Hence, the whole phenomena contains both the kinetic energy and the potential energy. The sum of kinetic energy and the potential energy is called Mechanical energy. Therefore, the correct answer is Mechanical Energy.</span>
8 0
3 years ago
Read 2 more answers
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
Read 2 more answers
A computer disk drive is turned on starting from rest and has a constant angular acceleration. If it took 0.750 seconds for the
Irina-Kira [14]

Answer:

a) Θ = ω₀*t + ½αt² To complete first revolution 2π rads = 0*t + ½αt² and to complete the first and second combined 4π rads = 0*t + ½α(t+0.810s)² Divide second by first: 2 = (t + 0.810s)² / t² This is quadratic in t and has roots at t = -0.336 s ← ignore and t = 1.96 s ◄ b) Use either equation from above: 2π rads = 0*t + ½α(1.96s)² α = 3.27 rad/s² ◄ Hope this helps!

Explanation:

6 0
3 years ago
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