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steposvetlana [31]
3 years ago
15

A 5.0 kg, 54-cm-diameter cylinder rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder i

s held with the center of mass at the same height as the axle, then released. (Figure 1)What is the magnitude of the cylinder's initial angular acceleration?What is the magnitude of the cylinder's angular velocity when it is directly below the axle?
Physics
1 answer:
bagirrra123 [75]3 years ago
8 0

Answer: a) 24.2 rad/s² b) 6.95 rad/s

Explanation:

So the cylinder's moment of inertia is I = \frac{MR^2}{2}, where M is the mass of the cylinder and R is its radius.

The problem you have here is that the axle is modifying it, so we use the parallel axis theorem:

Therefore

I =  \frac{MR^2}{2} + MD^2 where D is the distance from the center of mass to the axis which is a radius R

\=> I =  \frac{MR^2}{2} + M(R)^2 = \frac{3MR^2}{2}

Next the Momentum with respect to the center of mass of the cylinder is

M_{c} = I\alpha = F*R = m*g*R because there are no other forces rather than the cylinder's weight, now I = \frac{3*5*0.27^2}{2} = 0.54675 kg*m^2

a ) replace and find \alpha = (m*g*R)/(I )

\alpha = (5.0*9.8*0.27)/(0.54675 ) = 24.2 rad/s^2

b) we start by assuming that the energy is conserved because the cylinder is rolling without sliding:

so the potential energy is given by E = M*g*R = 5*9.8*0.27 = 13.23J

now you have that the same amount is going to be equal to the kinetic rotational energy of the cylinder so 13.23J = \frac{I*\omega^2}{2}  => \omega = \sqrt(\frac{2*13.23J}{0.54675} ) = 6.95 rad/s

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A number of compounds containing the heavier noble gases, and especially xenon, have been prepared. One of these is xenon hexafl
aalyn [17]

Answer:

The answer is "82.2 torr"

Explanation:

moles of Xe:

= \frac{0.06}{131.293} \\\\ =0.00045699313 \ \mol

moles of F_2:

= \frac{0.0274}{38} \\\\= 0.00072105263\ \  mol

moles of produced XeF_2:

= 0.00024

moles of left Xe:

= 0.00021

Calculate the Pressure:

= \frac{(0.08206\times 0.00024 \times 293)}{(.1)}   + \frac{(0.08206\times  0.00021 \times  293)}{(.1)} \\\\= 0.10819611 \ \ atm \\\\ = 0.10819611 \times  760 \\\\ = 82.2 \ \ torr

5 0
3 years ago
24. A total of 165 J of work is done on a gaseous refrigerant as it undergoes compression. If the internal energy of the gas inc
Yuliya22 [10]

The total amount of energy transferred as heat is equal to 288 Joules.

<u>Given the following data:</u>

  • Internal energy = 123 Joules
  • Work done = 165 Joules

To calculate the total amount of energy transferred as heat, we would apply the first law of thermodynamics.​

<h3>The first law of thermodynamics.</h3>

Mathematically, the first law of thermodynamics is given by the formula:

\Delta E = Q - W

<u>Where;</u>

  • \Delta E is the change in internal energy.
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Substituting the given parameters into the formula, we have;

123 = Q - 165\\\\Q=123+165

Q = 288 Joules.

Read more on internal energy here: brainly.com/question/25737117

4 0
3 years ago
A child's toy consists of a spherical object of mass 50 g attached to a spring. One end of the spring is fixed to the side of th
Artemon [7]

Answer:

0.14 J

Explanation:

The maximum velocity is the amplitude times the angular frequency.

vmax = Aω

ω = vmax / A

ω = (3.2 m/s) / (0.06 m)

ω = 53.3 rad/s

For a spring-mass system:

ω = √(k / m)

ω² = k / m

k = ω²m

k = (53.3 rad/s)² (0.050 kg)

k = 142 N/m

The elastic potential energy is:

EE = ½ kx²

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6 0
3 years ago
I think these two are simple questions.. but I need help asap..... TT
frosja888 [35]

1) The average velocity is 56 m/min

2) The average velocity is -83 m/min

Explanation:

1)

The average velocity of an object is given by

v=\frac{d}{t}

where

d is the displacement (change in position)

t is the time elapsed

In the graph in the problem, the displacement corresponds to the distance, therefore to the change in the y-variable (\Delta y), while the time elapsed is the change in the x-variable (\Delta x), so the average velocity can be written as

v=\frac{\Delta y}{\Delta x}

At point A, we have:

y_A = 5 m\\x_A = 0.1 min

At point B, we have:

y_E = 55 m\\x_A = 1 min

So, we have

\Delta y= 55 -5 = 50 m\\\Delta x = 1.0-0.1 = 0.9 min

So the average velocity is

v=\frac{50 m}{0.9 min}=56 m/min

2)

In this part instead, we have the following:

At point F, we have:

y_F = 55 m\\x_A = 1.3 min

At point H, we have:

y_H = 30 m\\x_A = 1.6 min

So, we have

\Delta y= 30 -55 = -25 m\\\Delta x = 1.6-1.3 = 0.3 min

So the average velocity is

v=\frac{-25 m}{0.3 min}=-83 m/min

Learn more about velocity:

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