Physics, 05.11.2019 02:31 ashleymartinez147

If you had a single pure mystery substance and cut it half over and over again until you eventually had two other different mystery substances the original substance would have been a(n) a. mixture. b. element. c. molecule. d. compound.

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Answer from: tiaharris3191

The correct answer is A. The original substance would have been a mixture.

Explanation:

When two or more substances (elements or compounds) come together but don't combine to make a new substance, a mixture results. The proportions of the substances in the mixture can be changed without changing the identity of the mixture.

If you had a single pure mystery substance and we obtained two other different mystery substances, that means that the original substance was a mixture of two other different mystery substances.

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

Answer from: Quest

you'd use the equation that establishes a relationship between energy and wavelength, which looks like this

e=h⋅cλ, where

λ - the wavelength of the photon; c - the speed of light, usually given as 3⋅108m/s; h - planck's constant, equal to 6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m

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