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kkurt [141]
3 years ago
5

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.77

7 m and 2.67 kg, respectively. When the propellor rotates at 573 rpm (revolutions per minute), what is its rotational kinetic energy?
Physics
1 answer:
Vikki [24]3 years ago
6 0
The formula for the rotational kinetic energy is

KE_{rot} = \frac{1}{2}(number \ of\ propellers)( I)( omega)^{2}

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

I=m L^{2}=(2.67 \ kg) (0.777 \ m)^{2}  =2.07459 \ kgm^{2}

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

KE_{rot} =( \frac{1}{2} )(5)(2.07459 \ kgm^{2}) (60\ rad/s)^{2}

KE_{rot} =18,671.31 \ J

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Answer: Take your pick

Explanation:

if they are all in parallel 1 /(1/100 + 1/300 + 1/50) = 30 Ω

if 50 is in parallel with 2 in series 1 / (1/(100 + 300) + 1/50) = 44.444...Ω

if 100 is in parallel with 2 in series 1 / (1/(50 + 300) + 1/100) = 77.777...Ω

if 300 is in parallel with 2 in series 1 / (1/(100 + 50) + 1/300) = 100 Ω

If 50 is in series with 2 in parallel 50 + 1/(1/100 + 1/300) = 125 Ω

If 100 is in series with 2 in parallel 100 + 1/(1/50 + 1/300) = 142.857...Ω

If 300 is in series with 2 in parallel 300 + 1/(1/50 + 1/100) = 333.333...Ω

If they are all in series 100 + 300 + 50 = 450 Ω

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2 years ago
Approximately where is it currently high tide on Earth? Group of answer choices wherever it is currently noon anywhere that ocea
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Option D, only on the portion of the Earth facing directly toward the Moon

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If you increase the mass but keep force constant what will happen to the acceleration of the object
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3 0
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Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.
Ksivusya [100]

|v| =\sqrt{ G \cdot M / r}, where

  • M the mass of the planet, and
  • G the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

<em>Equation 1</em>  (see below) relates net force the object experiences, \Sigma F to its orbit velocity v and its mass m required for it to stay in orbit :

\Sigma F = m \cdot v^{2} / r <em>(equation 1)</em>

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force \Sigma F acting on the soccer ball shall equal to its weight, W = m \cdot g where g the gravitational acceleration constant. Thus

\Sigma F = W = m \cdot g <em>(equation 2)</em>

Substitute equation 2 to the left hand side of <em>equation 1</em> and solve for v; note how the mass of the soccer ball, m, cancels out:

m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0) <em>(equation 3)</em>

<em>Equation 4 </em> gives the value of gravitational acceleration, g, a point of negligible mass experiences at a distance r from a planet of mass M (assuming no other stellar object were present)

g = G \cdot M/ r^{2} <em>(equation 4)</em>

where the universal gravitation <em>constant</em> G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}

Thus

\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}

3 0
3 years ago
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