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3 years ago

9

A pulley is able to lift a mass of 25 kg 0.30 m with an applied force of 50 N over a distance of 1.5 m. What is the ideal mechan

ical advantage of the pulley system?

1 answer:

spin [16.1K]3 years ago

7 0

The ideal mechanical advantage of the pulley system is 3

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Which of these is not a type of force?

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Inertia
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Two loudspeakers emit 600 Hz Hz notes. One speaker sits on the ground. The other speaker is in the back of a pickup truck. You h

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Answer:

The truck's speed is 4.04 m/s.

Explanation:

Given that,

Emit frequency = 600 Hz

Beat = 7.00 beat/sec

We need to calculate the truck's speed

Using formula of speed

$\text{frequency observed}=\text{frequency emitted}\times\dfrac{v}{v+v_{source}}$

Where, v = speed of sound

Put the value into the formula

$(600-7)=600\times(\dfrac{343}{343-v_{truck}})$

$v_{truck}=\dfrac{600\times343-593\times343}{593}$

$v_{truck}=4.04\ m/s$

Hence, The truck's speed is 4.04 m/s.

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3 years ago

A ball accelerates uniformly at +4.0 m/s2

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2 years ago

How fast was a plane flying if it traveled 500 km in 30 min?

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A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$

Finally, using the Parallel Axis Theorem, we calculate I_B:

$I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}$

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