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3 years ago

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A textbook of mass 2.02 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose d

iameter is 0.190 m , to a hanging book with mass 3.05 kg . The system is released from rest, and the books are observed to move a distance 1.19 m over a time interval of 0.800 s .
A) What is the tension in the part of the cord attached to the textbook?
B) What is the tension in the part of the cord attached to the book?
C) What is the moment of inertia of the pulley about its rotation axis?

1 answer:

Anika [276]3 years ago

7 0

Answer:

a. 7.38 N b. 40.87 N c. 0.113 kg-m²

Explanation:

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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

a

The radial acceleration is $a_c = 0.9574 m/s^2$

b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

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2 years ago

A 55.0-g sample of hot metal initially at 99.5oC was added to 40.0 g of water in a Styrofoam coffee cup calorimeter. The water a

Kaylis [27]

Answer:

Cp= 0.44 J/g.C

This is heat capacity of metal.

Explanation:

From energy conservation

Heat lost by metal = Heat gain by water +Heat gain by calorimeter

Because here temperature of metal is high that is why it loose the heat.The temperature of water and calorimeter is low that is why they gain the heat.

final temperature is T= 30.5 C

We know that sensible heat transfer given as

Q= m Cp ΔT

m=Mass

Cp=Specific heat capacity

ΔT=Temperature difference

By putting the values

55 x Cp ( 99.5 - 30.5) = 40 x 4.184 ( 30.5- 21 ) + 10 x ( 30.5 - 21)

Cp ( 99 .5- 30.5) = 30.65

Cp= 0.44 J/g.C

This is heat capacity of metal.

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3 years ago

Do atoms have full outer energy level combine with other atoms

Pavlova-9 [17]

All of the Noble Gases, which are on the right side of the periodic table, have a full outer energy level. The elements that are Noble Gases are the following: <span>Neon Argon Krypton Xenon Radon Ununoctium.

Hope this helps.</span>

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2 years ago

You are investigating how objects move when they are dropped from different heights. To collect your data, you drop a 1 kg weigh

ASHA 777 [7]

The time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

The given parameters;

<em>Mass of the first object, m1 = 1 kg</em>

<em>Mass of the second object, m2 = 5 kg</em>

The final velocity of the objects during the downward motion is calculated as follows;

$v_f = v_0 + gt\\\\v_f = 0 + gt\\\\\v_f = gt$

The time of motion of the object from the given height is calculated as;

$h = v_0 t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} }$

The time of motion of each object is independent of mass of the object.

Thus, the time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

Learn more about time of motion here: brainly.com/question/2364404

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2 years ago

A 3-kg mass is in free fall. What is the velocity of the mass after 7 seconds??

Lisa [10]

68.6m/s is the answer <span />

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3 years ago

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