Answer:
At center of ruler.
Explanation:
The center of gravity is defined as the point at which the object can be balanced. Due to different structure, weight and size, center of gravity changes from object to object.
In this case we have a plastic ruler which has a homogeneous structure, that means its weight is equally distributed through out. So its center of gravity must be at a point which is at equal distance from the both ends ( center ).
27) The connections between neurons in the retina, specifically the connections referred to as “lateral inhibition,” help us see which of the following better?
Contrast
Faces
Colors
45) Improving the contrast of an image (making the dark regions darker and the light regions lighter) helps us to identify:
The edges of objects
The center of objects
The color of an object
67What assumption does our visual system make in order to see curved surfaces (domes, holes)?
Light comes from above
Curved surfaces are always evenly lit
Curved surfaces are always easy to see, no assumptions are made
78) Which part of the face does our brain pay the most attention to?
Eyes and mouth
Eyes and ears
Eyes and chin
88) If all these assumptions sometimes lead to mistakes, for example in these optical illusions, why do we make them?
It helps us see things faster
It helps us see things correctly
It helps us pay attention to what's important
All of the above
To find the planet's radius in terms of the radius Rg of Earth, use the equation g = GM/R^2 and substitute 2g for g. Solve for R to get R = sqrt(1/(2gMg)) * Rg.
To find the planet's radius in terms of the radius Rg of Earth, we need to understand the relationship between the gravitational field and the mass and radius of a planet. The magnitude of the gravitational field on the surface of a planet is given by g = GM/R2, where G is the gravitational constant, M is the mass of the planet, and R is its radius. For the planet in question, we are told that the magnitude of the gravitational field is 2g and its mass is half the mass of Earth. Since the gravitational field is 2g, we can substitute g with 2g in the equation and solve for R in terms of Rg:
2g = GM/R2 → 2gR2 = GM → 2gR2 = (GMg)/(2Rg) → R2/Rg = 1/(2gMg) → R = sqrt(1/(2gMg)) * Rg
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To find the radius of a planet with a gravitational field twice that of Earth's and half the mass, the radius is calculated to be half of Earth's radius.
The magnitude of the gravitational field strength g on a planet is given by the equation g = G(M/R^2), where G is the universal gravitation constant, M is the planet's mass, and R is the planet's radius. Given that the gravitational field on the surface of the particular planet is 2g where g is Earth's gravitational field, and the planet's mass is half of Earth's mass, we can derive the planet's radius in terms of Earth's radius Rg. Setting up the proportion (G(1/2M_Earth)/(R^2)) / (G(M_Earth)/(Rg^2)) = 2, and simplifying, we find that R^2 = (1/4)Rg^2. Taking the square root of both sides gives us the final relation R = (1/2)Rg.
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Violet light frequency is 7.5 x 10^14 hertz. As according to the relation between light's speed, frequency and wavelength, λ = c / f, the wavelength of violet light can be found to be approximately 400 nanometers.
The relationship between frequency (f), wavelength (λ), and the speed of light (c) is expressed by the equation c = fλ. Given the speed of light c is approximately 3.00 × 10⁸ meters per second (m/s) and the frequency of violet light f is given as 7.5 × 10¹⁴ hertz (Hz), we can solve for wavelength λ. Rearranging the equation gives us λ = c / f. Substituting the given values we find that λ = (3.00 × 10⁸ m/s) / (7.5 × 10¹⁴ Hz), which computes to approximately 400 nanometers (nm). This result corroborates that violet light is typically within the range of 380 - 450 nm, thus confirming its position at the high-frequency, short-wavelength end of the visible light spectrum.
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