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3 years ago

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An escalator carries you from one level to the next in an airport terminal. The upper level is 4.4m above the lower level, and t

he length of the escalator is 7.3m . Assume that your mass is 63kg .Part AHow much work does the up escalator do on you when you ride it from the lower level to the upper level?Express your answer to two significant figures and include the appropriate units.W =___________.Part BHow much work does the down escalator do on you when you ride it from the upper level to the lower level?Express your answer to two significant figures and include the appropriate units.W =___________.

1 answer:

horrorfan [7]3 years ago

6 0

Answer:

Part a)

$W = 2716.5 J$

Part B)

$W = -2716.5 J$

Explanation:

Part A)

While escalator is moving up work done to move the person upwards is given as

$W = mgh$

here we know that

m = 63 kg

h = 4.4 m

now we have

$W = 63 \times 9.8 \times 4.4$

$W = 2716.5 J$

Part B)

Work done while we move from up to down

So we have

$W = - mgh$

so we have

$W = -2716.5 J$

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2 years ago

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A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

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Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

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$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

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3 years ago

child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s

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Answer:

θ = 13.7º

Explanation:

According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

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ΔK, is the change in kinetic energy, as follows:

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ΔU, is the change in the gravitational potential energy.
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Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

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Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

$\frac{1}{2}* (v_{f} ^{2} - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d$

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