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3 months ago

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9. A 2 liter bottle of Coke weighs about 2 kilograms. If that bottle were combined with an equal amount of anti-Coke, how many J

oules of energy would be released (note that you get units Joules if you calculate the energy using units of kilograms and meters)

1 answer:

Maru [3.3K]3 months ago

3 0

Answer:

The energy expected to be released is calculated to be 4182 Joules.

Explanation:

The total mass of coke is 2 kg, which is equivalent to 2000 g

1 calorie per gram corresponds to 4.184 Joules of energy

4.184 J/gC * 2000g results in 8368 J

1 food calorie approximates to 4186 J

By subtracting, we find 8368 - 4186

Hence, the total energy that will be released amounts to 4182 Joules.

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The figure above represents a stick of uniform density that is attached to a pivot at the right end and has equally spaced marks

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Answer:

After a duration of 2.0 seconds, the angular momentum of the system is $L= 2(4A+3B+2C+D)x$ .

Explanation:

Let’s denote the forces acting on the rod as A, B, C, and D, and the distance between them as $x$ .

The torque produced by force A can be expressed as

$\tau_a = 4Ax$ ,

for force B the torque is

$\tau_b = 3Bx$ ,

the torque for force C is

$\tau_c = 2Cx$ ,

and the torque due to force D is

$\tau_d = Dx$ .

This leads to a total torque on the stick as

$\tau_{tot} =\tau_a+\tau_b+\tau_c+\tau_d$

$\tau_{tot} =4Ax+3Bx+2Cx+Dx$

$\tau_{tot} =x(4A+3B+2C+D)$

Subsequently, this torque results in an angular acceleration $\alpha$ in accordance with the formula

$I \alpha = \tau_{tot}$

where $I$ represents the moment of inertia of the stick, which has a value of

$I = \dfrac{1}{3} m(4x)^2$ .

Thus, the angular acceleration calculates to

$\alpha = \dfrac{\tau_{tot} }{I}$

$\alpha =\dfrac{x(4A+3B+2C+D)}{\dfrac{1}{3}m(4x)^2 }$

$\boxed{\alpha =\dfrac{3(4A+3B+2C+D)}{16mx } .}$ .

Now, the angular momentum $L$ of the rod is given by

$L = I\omega$ ,

where $\omega$ is considered the angular velocity.

Given $\omega = \alpha t$ , we conclude that

$L = \dfrac{1}{3}m (4x)^2 *\dfrac{3(4A+3B+2C+D)}{16mx }* t$

$L= (4A+3B+2C+D)x t$

Thus, $t = 2.0s$ , the angular momentum is

$\boxed{ L= 2(4A+3B+2C+D)x. }$

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