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3 years ago

"A window washer pulls herself upward using the bucket-pulley apparatus. The mass of the person plus the bucket is 65 kg.

Physics

2 answers:

FromTheMoon [43]3 years ago

8 0

Part A.

If there is a constant speed, therefore the net force is zero.

Forces acting are the:

1. Weight

2. Upward force exerted by the rope which is opposite in direction but the magnitude is to the force which is acting downwards exerted on the rope.

∑Force = pulling force - weight = 0

Weight = mg

Hence,

F (pulling force) = mg

Substituting the values,

F (pulling force) = 65 × 9.81

F (pulling force) = 637.65 Newton

____________________________________________________________

Part B.

F = ma

10% increase in the force = 1.10 ₓ 637.65

10% increase in the force = 701.415

Therefore,

701.415 = 65 ₓ a

a = 701.45/65

a = 10.8 m/s²

Aleksandr [31]3 years ago

7 0

Consider the free body diagrams as shown in the file attached.

Let the acc of system is a. <span>
For the washer T+N-m1g=m1a ............................(1)
For the bucket T-N-m2g=m2a ............................(2)
</span>
(A)
a=0 Eqns (1) and (2) become T+N-m1g=0 T-N-m2g=0 Adding and rearranging, T=(m1+m2)/2g T=325N <span>
(B)
</span> We are specified the value of T here as 325+32.5=357.5N eqns (1)and (2) become 357.5+N-m1g=m1a 357.5-N-m2g=m2a Adding, 715-(m1+m2)g=(m1+m2)a

Solve for 'a'

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2 years ago

2. A cinder block is sitting on a platform 20 m high. It weighs 16kg. The block has

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Answer:

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Explanation:

Applying,

P.E = mgh.............. Equation 1

Where P.E = potential energy, m = mass of the cinder block, h = height of the platform, g = acceleration due to gravity.

From the question,

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P.E = 16(20)(9.8)

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2 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

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3 years ago

28. Which of the following correctly shows the order of highest amount of friction to the lowest amount of

Bad White [126]

Answer:

$\mathrm{d.\:Static,\: Sliding,\:Rolling}$

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Static friction occurs when an object initially starts at rest. When the surfaces of the materials touch, the microscopic unevenness interlock greatest with each other, causing the most friction out of the three.

During sliding friction, an object is already moving or in motion. The microscopic surfaces still interlock, but because the object is in motion, it has a momentum. Therefore, the magnitude of sliding friction is less than that of static friction.

Rolling friction occurs when an object rolls across some surface. Rather than surfaces interlocking, rolling friction is caused by the constant distortion of surfaces. As it rolls, the surfaces of the object are constantly wrapping and changing. This distortion causes the rolling friction. However, it is much less in magnitude when compared to static or sliding friction.

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2 years ago

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Thus, the image that shows these layers of graphite is attached to this an answer

Learn more about graphite:brainly.com/question/11095487

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1 year ago