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Licemer1 [7]
3 years ago
7

A massless rod of length L has a small mass m fastened at its center and another mass m fastened at one end. On the opposite end

from the mass m, the rod is hinged with a frictionless hinge. The rod is released from rest from an initial horizontal position; then it swings down. What is the angular velocity as the rod swings through its lowest (vertical) position?
Physics
1 answer:
konstantin123 [22]3 years ago
3 0

Answer:

onservation of energy

U top = K bottom

(m + m)*g*L = 1/2*I*?^2 where I = m*(L/2)^2 + m*L^2 = 1.25*m*L^2

So 2m*g*L = 1/2*1.25*m*L^2*?^2

So ? = sqrt(3*g*/(1.25*L) ) = sqrt(12g/5L)

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Answer:

W = 19.845 J

Explanation:

Work is defined as W = Fdcos\theta, where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself, \theta = 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:

W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m

W = 19.845 J

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De que trata la primera ley de newton
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uhm i think i don’t understand what it says but i can answer if it was English ! i also tried to translate it but it didn’t help !
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How do i solve the ones in red? A student fires a cannonball horizontally with a speed of 24.0m/s from a height of 51.0m. Neglec
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Horizontal speed = 24.0 m/s

height of the cliff = 51.0 m

For the initial vertical speed will are considering the vertical component. Therefore,

Since the student fires the canonical ball at the maximum height of 51 m, the initial vertical velocity will be zero. This means

\text{ Initial vertical velocity = 0 m/s}

let's find how long the ball remained in the air.

\begin{gathered} 0=51-\frac{1}{2}(9.8)t^2 \\ 4.9t^2=51 \\ t^2=\frac{51}{4.9} \\ t^2=10.4081632653 \\ t=\sqrt[]{10.4081632653} \\ t=3.22 \\ t=3.22\text{ s} \end{gathered}

Finally, let's find the how far from the base of the building the ball landed(horizontal distance)

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