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Perform calculations using the circuit illustrated. Round

Physics

1 answer:

balu736 [363]3 years ago

6 0

Answer:5.9, 1.0. D

Explanation:

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The drawing shows two frictionless inclines that begin at ground level (h = 0 m) and slope upward at the same angle θ. One track

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Answer:

a). $H=2.45m$

b). $H_{max}=1.94m$

Explanation:

For the block that stays on the track, its maximal height is attained when all of the kinetic energy is converted to potential energy

a).

The height for the block on the longer track can by find using this equation:

$\frac{1}{2}*m*v_o^2=m*g*H$

Cancel the mass as a factor in each element in the equation

$H=\frac{v_o^2}{2*g}$

$H=\frac{(6.94m/s)^2}{2*9.8m/s^2}$

$H=2.45m$

b).

The other lost some kinetic energy so, use a projectile motion to determine the total height for the other bock:

$E_k=E_p$

$E_k=m*g*H_1$

$E_k=\frac{1}{2}*m*v_o^2-\frac{1}{2}*m*v^2$

$m*g*H_1=\frac{1}{2}*m*(v_o^2-v^2)$

Solve to v'

$v^2=v_o^2-2*g*H_1$

$v=\sqrt{v_o^2-2*g*H_1}=\sqrt{(6.94m/s)^2-2*9.8m/s^2*1.25m}$

$v=4.8m/s$

$H_{max}=H_1+\frac{v^2*sin(50)}{2*g}=1.25m+\frac{(4.8m/s)^2*sin(50)}{2*9.8m/s^2}$

$H_{max}=1.94m$

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