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nadezda [96]
2 years ago
14

Cylinder A is moving downward with a velocity of 3 m/s when the brake is suddenly applied to the drum. Knowing that the cylinder

moves 6 m downward before coming to rest and assuming uniformly accelerated motion, determine (a) the angular acceleration of the drum, (b) the time required for the cylinder to come to rest.
Physics
1 answer:
Xelga [282]2 years ago
7 0

Answer:

Incomplete question

Check attachment for the given diagram

Explanation:

Given that,

Initial Velocity of drum

u=3m/s

Distance travelled before coming to rest is 6m

Since it comes to rest, then, the final velocity is 0m/s

v=3m/s

Using equation of motion to calculate the linear acceleration or tangential acceleration

v²=u²+2as

0²=3²+2×a×6

0=9+12a

12a=-9

Then, a=-9/12

a=-0.75m/s²

The negative sign shows that the cylinder is decelerating.

Then, a=0.75m/s²

So, using the relationship between linear acceleration and angular acceleration.

a=αr

Where

a is linear acceleration

α is angular acceleration

And r is radius

α=a/r

From the diagram r=250mm=0.25m

Then,

α=0.75/0.25

α =3rad/sec²

The angular acceleration is =3rad/s²

b. Time take to come to rest

Using equation of motion

v=u+at

0=3-0.75t

0.75t=3

Then, t=3/0.75

t=4 secs

The time take to come to rest is 4s

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Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors a
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Answer:

E=0

Explanation:

Electric field due to each thin sheet of charge=\sigma/2\varepsilon

let us say the right plate has positive charge density \varepsilonand left sheet has a negative charge density -\varepsilon .

In the region between the plates,the electric field due to each plate is in same direction,

E=\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=\sigma/\varepsilon

in the region outside the plates, the field due to the plates is in opposite directions

E=-\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=-\sigma/2\varepsilon+\sigma/2\varepsilon

E=0

4 0
2 years ago
What visible signs indicate a precipitation reaction when two solutions are mixed?
Illusion [34]

Formation of an insoluble solid

Explanation:

One of the remarkable visible signs that indicates a precipitation reaction when two solutions are mixed is the formation of an insoluble solid. The insoluble solid formed is the precipitate.

  • Precipitates usually forms in single replacement reactions and double replacement or double decomposition reactions.
  • They form when two soluble compounds react. One of the product is an insoluble solid in the solution called the precipitate.
  • The solubility table helps to predict whether precipitates forms in a reaction.

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6 0
2 years ago
If a satellite is orbiting the Earth in elliptical motion, then it will move _______________ (slowest, fastest) when its closest
AVprozaik [17]

Answer:fastest,same,slow down,opposite,slow

Explanation:

8 0
1 year ago
A 12 cm diameter piston-cylinder device contains air at a pressure of 100 kPa at 24oC. The piston is initially 20 cm from the ba
lina2011 [118]

Answer:

ΔQ = 0.1 kJ

\mathbf{v_f = 1.445*10^{-3}  m^3}

\mathbf{P_f = 156.5 \ kPa}

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

Explanation:

GIven that:

Diameter of the piston-cylinder = 12 cm

Pressure of the piston-cylinder = 100 kPa

Temperature =24 °C

Length of the piston = 20 cm

Boundary work ΔW = 0.1 kJ

The gas is compressed and The temperature of the gas remains constant during this process.

We are to find ;

a. How much heat was transferred to/from the gas?

According to the first law of thermodynamics ;

ΔQ = ΔU + ΔW

Given that the temperature of the gas remains constant during this process; the isothermal process at this condition ΔU = 0.

Now

ΔQ = ΔU + ΔW

ΔQ = 0 + 0.1 kJ

ΔQ = 0.1 kJ

Thus; the amount of heat that was transferred to/from the gas is : 0.1 kJ

b. What is the final volume and pressure in the cylinder?

In an isothermal process;

Workdone W = \int dW

W = \int pdV \\ \\ \\W = \int \dfrac{nRT}{V}dv \\ \\ \\ W = nRt \int  \dfrac{dv}{V}  \\ \\ \\ W  = nRT In V |^{V_f} __{V_i}}  \\ \\ \\ W = nRT \ In \dfrac{V_f}{V_i}

Since the gas is compressed ; then v_f< v_i

However;

W =- nRT \ In \dfrac{V_f}{V_i}

W =- P_1V_1  \ In \dfrac{V_f}{V_i}

The initial volume for the cylinder is calculated as ;

v_1 = \pi r^2 h \\ \\   v_1 = \pi r^2 L \\ \\ v_1 = 3.14*(6*10^{-2})^2*(20*10^{-2}) \\ \\ v_1 = 2.261*10^{-3} \ m^3

Replacing over values into the above equation; we have :

100 =  - ( 100*10^3 *2.261*10^{_3}) In (\dfrac{v_f}{v_i}) \\ \\ - In (\dfrac{v_f}{v_i})= \dfrac{100}{(100*10^3*2.261*10^{-3})} \\ \\ - In \ v_f  + In \  v_i = \dfrac{100}{226.1} \\ \\   - In \ v_f  = - In \ v_i + \dfrac{100}{226.1}  \\ \\  - In \ v_f  = - In (2.261*10^{-3} + \dfrac{100}{226.1 } \\ \\  - In \ v_f  = 6.1 + 0.44 \\ \\  - In \ v_f  = 6.54 \\ \\  - In \ v_f  = -6.54 \\ \\ v_f = e^{-6.54} \\ \\ \mathbf{v_f = 1.445*10^{-3}  m^3}

The final pressure can be calculated by using :

P_1V_1 = P_2V_2 \\ \\ P_iV_i = P_fV_f

P_f =\dfrac{P_iV_i}{V_f}

P_f =\dfrac{100*2.261*10^{-3}}{1.445*10^{-3}}

P_f = 1.565*10^2 \ kPa

\mathbf{P_f = 156.5 \ kPa}

c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?

The change in entropy of the gas is given  by the formula:

\Delta S=\dfrac{\Delta Q}{T}

where

T =  24 °C = (24+273)K

T = 297 K

\Delta S=\dfrac{-100 \ J}{297 \ K}

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

4 0
2 years ago
The gravitational force between the Sun (mass = 1.99 × 1030 kg) and Mercury (mass = 3.30 × 1023 kg) is 8.99 × 1021 N. How far is
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The answer is B - 6.98 x 10^7 km
8 0
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