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3 years ago

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A point P1 is located by the vector A = (3.74)î + (1.64)ĵ and a point P2 is located by the vector B = (1.60)î + (3.66)ĵ. The vec

tor C locates the point P2 relative to the point P1. Determine the following. (a) the vector Crightharpoonhead.gif which points from the point P1 to the point P2

1 answer:

seropon [69]3 years ago

4 0

Answer:

$\vec{C} = ( \ - 2.14 \ , \ 2.02 \ )$

Explanation:

The vector that point from point P1 to point P2 its found simply by taking the vector at which point P2 its located and subtracting the vector at which point P1 its located:

$\vec{C} = \vec{B} - \vec{A}$

So:

$\vec{C} = ( \ 1.60 \ , \ 3.66 \ ) - ( \ 3.74 \ , \ 1.64 \ )$

$\vec{C} = ( \ 1.60 \ - \ 3.74 \ , \ 3.66 \ - \ 1.64 \ )$

$\vec{C} = ( \ - 2.14 \ , \ 2.02 \ )$

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A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i

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Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

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spring constant, k = 250 N/m

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u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

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on substituting the values, we get

10 × 9.89 = 5 × v₂

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v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

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Work done by the friction force = change in kinetic energy of the mass 2

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or

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Energy of the spring = Kinetic energy of the mass 2

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Answer:

$8.8\times 10^{-6} W/m^2$

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$1 mm=10^{-3} m$

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$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

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$I=8.8\times 10^{-6} W/m^2$

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Explanation:

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