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lara [203]
3 years ago
12

You are creating a carbon cycle poster. What could you include in the poster to demonstrate the cycling

Physics
1 answer:
nalin [4]3 years ago
3 0

Answer:

the answer should be d.

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How much work is done if a force of 20N is used to move an object 6 metres? <br><br> pls help
Dmitriy789 [7]

I assume that the force of 20 N is applied along the direction of motion and was applied for the whole 6 meters, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula; Work = 20 N * 6 m * cos 0º.

Work = 20 N * 6 m * 1

Work = 120 Nm

Work = 120 joules

Hope this helps!

8 0
3 years ago
Read 2 more answers
According to the diagram, which color of visible light has the shortest wavelength?
Fynjy0 [20]

Answer:

Violet Light

Explanation:

On one end of the spectrum is red light, with the longest wavelength. Blue or violet light has the shortest wavelength. White light is a combination of all colors in the color spectrum. It has all the colors of the rainbow.

6 0
3 years ago
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d is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?
Anna71 [15]
<span>D is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?
</span>Answer: The sled's potential energy is 882 Joules
5 0
3 years ago
Read 2 more answers
A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.
Fed [463]
<h2>Option A is the correct answer.</h2>

Explanation:

Acceleration due to gravity

                  g=\frac{GM}{r^2}

         G = 6.67 × 10⁻¹¹ m² kg⁻¹ s⁻²

  Let mass of earth be M and radius of earth be r.

  We have

               g=\frac{GM}{r^2}

Now

         A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.

       Mass of hypothetical planet, M' = M/2

       Radius of hypothetical planet, r' = 2r

  Substituting

              g'=\frac{GM'}{r'^2}\\\\g'=\frac{G\times \frac{M}{2}}{(2r)^2}\\\\g'=\frac{\frac{GM}{r^2}}{8}\\\\g'=\frac{g}{8}

Option A is the correct answer.

6 0
3 years ago
A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn
inn [45]

Answer:

The answer to the question is

The ladybug begins to slide

Explanation:

To solve the question we assume that the frictional force of the ladybug and the gentleman bug are the same

Where the  frictional force equals F_{Friction} = μ×N = m×g×μ

and the centripetal force is given by m·ω²·r

If we denote the properties of the ladybug as 1 and that of the gentleman bug as 2, we have

m₁×g×μ = m₁·ω²·r₁ ⇒ g×μ = ω²·r₁

and for the gentleman bug we have

m₂×g×μ = m₂·ω²·r₂ ⇒ g×μ = ω²·r₂

But r₁ = 2×r₂

Therefore substituting the values of r₁ =2×r₂ we have

g×μ = ω²·r₁ = g×μ = ω²·2·r₂

Therefore   ω²·r₂ = 0.5×g×μ for the ladybug. That is the ladybug has to overcome half the frictional force experienced by the gentleman bug before it start to slide

The ladybug begins to slide

6 0
3 years ago
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