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3 years ago

How are engineers creating clothing that can charge your cell phone. Explain the science behind their innovation.

Physics

1 answer:

soldier1979 [14.2K]3 years ago

8 0

Answer:

By the use of carbon nanotubes as the clothing material.

Explanation:

Carbon nanotubes is a technology designed by engineers that could make charging of the cell of a phone possible. It has some some required properties like it is strong, conductive and heat-resistant which are suitable for the purpose.

This technology which involves a progressive dipping of a pure cotton material into nafion polymer, then in a chemical solution of carbon nanotubes and finally in polystyrene until it becomes saturated and have a conductive property. The conductive cotton can be used to make such cloths which becomes conductive and produces some electromagnetic waves when there is a contact with sweat or fluid from the body. The waves generated can be used to charge a cell phone through radiation interaction.

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Ne4ueva [31]

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so at highest point

$v_1 = v_x$

now when he is at half of the maximum height the speed will be in x and y direction both

$v_2 = \sqrt{v_y^2 + v_x^2}$

here it is given that

$v_1 = 0.58 v_2$

$v_x = 0.58\sqrt{v_x^2 + v_y^2}$

$2.97 v_x^2 = v_x^2 + v_y^2$

$1.97 v_x^2 = v_y^2$

also we know that

$v_y^2 = v_{iy}^2 - 2 g \frac{H}{2}$

here we know that maximum height is given as

$H = \frac{v_{iy}^2}{2g}$

$v_y^2 = v_{iy}^2 - 2 g\frac{v_{iy}^2}{4g}$

$v_y^2 = \frac{v_{iy}^2}{2}$

now from above

$1.97 v_x^2 = \frac{v_{iy}^2}{2}$

$1.98 v_x = v_{iy}$

also we know that angle of projection is

$tan\theta = \frac{v_{iy}}{v_x}$

$tan\theta = \frac{1.98v_x}{v_x}$

so angle is

$\theta = tan^{-1} 1.98$

$\theta = 63.3 degree$

6 0

3 years ago

2. If you are 5'10" tall, that is, 5 feet 10 inches, what is your height in meters? (2 54 cm = 1.00 in)

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Answer:D 1.7

Explanation:

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2 years ago

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Answer:

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2 years ago

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PSYCHO15rus [73]

Answer:

F=50kg

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2 years ago

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Contact [7]

The center of mass is given with this formula:
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$v=\frac{dv}{dt}$
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$\frac{dx_c}{dt}=\frac{\sum_{n=1}^{n=i}d(m_ix_i)}{Mdt}\\
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3 years ago