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4 years ago

15

At how many rpm must the disk turn to raise the elevator at 35.0 cm/s

1 answer:

k0ka [10]4 years ago

3 0

<span>Let R be the radius of the disk, in centimeters. 35.0 cm/s = 35.0 * 60 cm/min = 2100 cm/min. One revolution of the disk is 2*pi*R, in cm. So the number of revolutions per minute is 2100 / (2*pi*R) = 1050/(pi*R), approximately 334/R.</span>

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A car of mass m accelerates from rest along a level straight track, not at constant acceleration but with constant engine power,

lisabon 2012 [21]

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3 years ago

jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en

d1i1m1o1n [39]

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

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so the angular speed of the train is given as

$\omega = \frac{\theta}{t}$

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4 years ago

Express 4,560 m in Km

Gala2k [10]

4.56 km is the answer

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3 years ago

Read 2 more answers

A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

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