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2 years ago

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A block slides down a slope inclined at 37 degrees. If the slope is frictionless and the block starts off at rest, how fast is i

t moving after it has traveled 1 meter along the plane? Apply energy-conservation principles to solve.

1 answer:

lianna [129]2 years ago

3 0

Answer:

Vf= 3.435 m/s

Explanation:

Given:

Initial velocity Vi =0 m/s (starting from Rest position)

θ = 37⁰

Distance S = 1 m

To find: Final Velocity Vf=?

fist we have to find the down slope net acceleration a = g sin θ

a= 9.81 sin 37⁰ = 5.9 m/s²

By 3rd equation of motion

2 a S= Vf² - Vi²

Vf = Square root ( 2 × 5.9 m/s² × 1 + 0 m/s)

Vf = Square root (11.8)

Vf= 3.435 m/s

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For each object we have then,

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Since the momentum of object A is greater than that of object B, then object A will make you feel force upon impact.

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3 years ago

The equation for density is mass divided by volume. the mass of a substance is changed so that it is made of fewer particles. If

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Density is directly proportional to mass. So if there's less matter inside object, its density will also reduce.

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2 years ago

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mr_godi [17]

Answer and Explanation:

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Where

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R represents Rydberg's constant

$n_f$ represents Final energy states

and $n_i$ represents initial energy states

Now Substitute is

$1.097\times 10^7\ m^{-1}\ for\ R, \infty for\ n_i,\ 3 for\ n_i,\\\\\ \frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )$

now we will put the values into the above formula

$= 1.097\times 10^7 m^{-1}(\frac{1}{3^2} - \frac{1}{\infty^2} )\\\\ = 1.097\times10^7\ m^{-1} (\frac{1}{9} )$

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Now we will rewrite the answer in the term of $\lambda$

$\lambda = \frac{1}{1218888.889} m\\\\ = 0.82\times 10^{-6} m$

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3 years ago

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Answer:

$2.46\cdot 10^5 J$

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kodGreya [7K]

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