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3 years ago

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The cylindrical tub of a dryer in a laundromat rotates counterclockwise about a horizontal axis at 41.5 rev/min as it dries the

clothes. If the diameter of the cylinder is 0.748 m, at what angle will a piece of cloth lose contact with the wall of the cylinder and fall down? (Take the +x axis to be pointing horizontally to the right.) degree counterclockwise from the +x axis

1 answer:

frozen [14]3 years ago

6 0

Answer:

$\theta = 49.81^0$

Explanation:

Given that:

$\omega = 41.5 \ rev/min\\\\\omega = 41.5 *\frac{1}{60}* 2 \pi\\\\\omega = 4.45 \ rad/s\\\\\\diameter = 0.748 m$

If we let the piece of the close lose contact at ∠θ;

Then ; from force balance;

we have:

$\\\\mg sin \theta = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})$

where;

$v = \frac{\omega d}{2}\\\\v = \frac{4.45 *0.748}{2}\\\\v = 1.6643\\\\v^2 = 2.77$

Again:

$\theta = sin^{-1}(\frac{2v^2}{dg})\\\\\theta = sin^{-1}( \frac{2*2.77}{0.74*9.8})\\\\\theta = 49.81^0$

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A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.9 m/s. The car is a distanc

sammy [17]

If it takes $t$ seconds to reach the car, then the distance $d$ is $3.9t$ .

The bear's distance from the tourist's starting point is
$6t-23$

For maximum $d$ , we set the equations equal to each other:
$3.9t=6t-23$
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3 years ago

When you drop a 0.43 kg apple, Earth exerts

Archy [21]

Answer:

The acceleration of the earth is 7.05 * 10^-25 m/s²

Explanation:

<u>Step 1:</u> Data given

mass of the apple = 0.43 kg

acceleration = 9.8 m/s²

mass of earth = 5.98 * 10 ^24 kg

<u>Step 2:</u> Calculate the acceleration of the earth

Following the third law of Newton F = m*a

F(apple) = F(earth) = m(apple)*a(apple)

F(apple) = 0.43 kg * 9.8 m/s² = 4.214 N

a(earth) = F(apple/earth)/m(earth)

a(earth) = 4.214N /5.98 * 10 ^24 kg

a(earth) = 7.05 * 10^-25 m/s²

The acceleration of the earth is 7.05 * 10^-25 m/s²

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3 years ago

(a) An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kin

Iteru [2.4K]

Answer:

A) The smaller object; B) The larger object

Explanation:

A) Lets say the small object is 2 kg and the large one is 6 kg. Lets say they also have 30 kg*m/s of momentum each. The small object would have 15 m/s velocity and the large would have 5 m/s.

Now for kinetic energy(.5*m*v²), the small object is .5*2*15², which is 225 J

The large object is .5*6*5², which is 75 J, so the smaller object has more Kinetic energy. Since velocity is squared, it is more important than how large mass is.

B) Same masses as part A. Lets say the kinetic energy is 45 J for both of them. For the small object, 45=.5*2*v²

.5*2 is 1, so 45/1 is 45. Take the square root and we get v= 6.71 m/s

For the large object, 45=.5*6*v²

.5*6 is 3, so 45/3 is 15. Take the square root and we get v=3.87 m/s

Now we plug the velocities into p=mv

For the small object, p=2*6.71, which gives us p=13.42 kg*m/s

For the large object, p=6*3.87, which gives us p=23.22 kg*m/s

The larger object has the larger momentum.

Hope this helps

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3 years ago

What is one way to describe elements

Talja [164]

Answer:

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3 years ago

An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

8 0

3 years ago