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3 years ago

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A 1.53-kg bucket hangs on a rope wrapped around a pulley of mass 7.07 kg and radius 66 cm. This pulley is frictionless in its ax

le and has the shape of a solid uniform disk. After the bucket has been released, what is the angular acceleration of the pulley

1 answer:

levacccp [35]3 years ago

7 0

Answer:

$\alpha = 6.431\,\frac{rad}{s^{2}}$

Explanation:

The pulley is modelled by the Newton's Laws, whose equation of equilibrium is:

$\Sigma M = T \cdot R = \frac{1}{2}\cdot M \cdot R^{2}\cdot \alpha$

Given that tension is equal to the weight of the bucket, the angular acceleration experimented by the pulley is:

$T = \frac{1}{2}\cdot M \cdot R \cdot \alpha$

$m_{b}\cdot g = \frac{1}{2}\cdot M \cdot R \cdot \alpha$

$\alpha = \frac{2\cdot m_{b}\cdot g}{M\cdot R}$

$\alpha = \frac{2\cdot (1.53\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{(7.07\,kg)\cdot (0.66\,m)}$

$\alpha = 6.431\,\frac{rad}{s^{2}}$

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What can you say about the magnitudes of the forces that the balloons exert on each other?

maxonik [38]

Answer:

$F_G=G. \frac{m_1.m_2}{R^2}$ gravitational force

$F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}$ electrostatic force

Explanation:

The forces that balloons may exert on each other can be gravitational pull due to the mass of the balloon membrane and the mass of the gas contained in each. This force is inversely proportional to the square of the radial distance between their center of masses.

The Mutual force of gravitational pull that they exert on each other can be given as:

$F_G=G. \frac{m_1.m_2}{R^2}$

where:

$G=$ gravitational constant $=6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}$

$m_1\ \&\ m_2$ are the masses of individual balloons

$R=$ the radial distance between the center of masses of the balloons.

But when there are charges on the balloons, the electrostatic force comes into act which is governed by Coulomb's law.

Given as:

$F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}$

where:

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$q_1\ \&\ q_2$ are the charges on the individual balloons

R = radial distance between the charges.

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3 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

<h3>a.</h3>

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

<h3>b.</h3>

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

<h2>b</h2>

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

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What is the use of 'ground' in electric circuit?

ss7ja [257]

In electrical engineering, ground or earth is the reference point in an electrical circuit from which voltages are measured, a common return path for electric current, or a direct physical connection to the earth. Electrical circuits may be connected to ground (earth) for several reasons.

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The windshield of a speeding car hits a hovering insect. Consider the time interval from just before the car hits the insect to

HACTEHA [7]

Answer:

A. False

B True

C. False

D.False

E. True

F. False

G. False

H. False

I. True

Explanation:

A. False: The system being analyzed consists of the bug and the car. These are the two bodies involved in the collision.

B. True: The system being analyzed consists of the bug and the car

C. False: The magnitudes of the change in velocity are different from the car and the bug. The velocity of the bug changes from 0 to the velocity of the car, while there is no noticeable change in the velocity of the car

D.False: There is barely any change in the momentum of the car since the mass of the bug is very small.

E. True: Since the mass of the bug is small, and was initially at rest, the magnitude of the change in monentum will be large because the new velocity will be that of the car.

F. False: The system being analyzed consists of the bug and the car. Those are the two bodies involved in the collision

G. False: The car barely changes in velocity since the mass of the bug is small.

H. False: The car barely changes in momentum because the collision does not affect its speed so much. on the other hand the momentum change of the bug is large since its mass is small.

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3 years ago

Which expression is equivalent to g−m ÷ gn?

Vinvika [58]

<u>Answer</u>

(g²n - m)/(gm)

<u>Explanation</u>

g - m ÷ gn = g - m/gn

Make the equation have the same denominator

g - m ÷ gn = g - m/gn = (ggn)/gn - m/gn

= (g²n)/gn - m/gm

Since they have the same denominator, we can carry out the subtraction on the numerator and then put them under one denominator.

(g²n)/gn - m/gm = (g²n - m)/(gm)

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3 years ago