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Tanzania [10]
3 years ago
9

Which would be most useful for overcoming the disadvantage of using ores? using public transportation recycling aluminum product

s buying paper instead of plastic minimizing electricity usage 100 points
Physics
2 answers:
nalin [4]3 years ago
7 0
The appropriate response is recycling aluminum products. Metal is a store in the Earths outside layer of at least one important minerals. The most profitable metal stores contain metals pivotal to industry and exchange, similar to copper, gold, and iron. Copper mineral is dug for an assortment of modern employments. Copper, an incredible conduit of power, is utilized as electrical wire.
Lady bird [3.3K]3 years ago
5 0

Answer: recycling aluminium products

Ore is an aggregate of inorganic minerals, which are economically valuable. The ores are present in the rocks. The ores are extracted from earth through the process of mining. The minerals present in the ores are non-metals and metals.

Recycling of aluminium products is most useful for overcoming the disadvantage of using ores, because ores are non-renewable resources, which means that their abundance is low on earth and these resources once used cannot be replenish. Aluminium is a valuable metal extracted from ore, therefore, once used for a specific purpose as a product should be recycled in another form so that can be used again.  

 

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It's either 3 or 4 I know this becuase I have read a book about electricity
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Answer:

the distance that the object is raised above its initial position is 5.625 m.​

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \  by \ the \ effort}{disatnce \ moved \  by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m

Therefore, the distance that the object is raised above its initial position is 5.625 m.​

3 0
2 years ago
The attraction will vary directly with the separation between the charges.
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3 years ago
A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th
Evgen [1.6K]

Answer:

\dfrac{d\theta}{dt}=0.038\ rad/s

Explanation:

Given that

\dfrac{dx}{dt}= -1\ m/s

From the diagram

tan\theta=\dfrac{21}{x}

By differentiating with time t

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

When x= 10 m

tan\theta=\dfrac{21}{10}

θ = 64.53°

Now by putting the value in equation

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)

\dfrac{d\theta}{dt}=0.038\ rad/s

Therefore rate of change in the angle is 0.038\ rad/s

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Answer:

c

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