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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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Energy that is transferred from a warmer object to a cooler object is called
alukav5142 [94]

Answer:

Heat

Explanation:

Energy that is transferred from a warmer object to a cooler object is called heat.

3 0
2 years ago
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Rays of light strike a bumpy road and are reflected.
ser-zykov [4K]

Answer:

last option is the correct one

6 0
3 years ago
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An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl
svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 0 m/s

         Displacement, s = 192 m

         Acceleration, a = 9.81 m/s²

         Substituting

                      s = ut + 0.5 at²

                      192 = 0 x t + 0.5 x 9.81 x t²

                         t = 6.26 seconds

         Now we need to find horizontal distance traveled in 6.26 seconds by the package.

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 41 m/s

        Time, t = 6.26 s

         Acceleration, a = 0 m/s²

         Substituting

                      s = ut + 0.5 at²

                      s = 41 x 6.26 + 0.5 x 0 x 6.26²

                         s = 256.52 m

     The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

          Initial velocity, u = 0 m/s

         Time, t = 6.26 s

         Acceleration, a = 9.81 m/s²  

         Substituting

                      v = u+ at

                       v = 0 + 9.81 x 6.26 = 61.41 m/s

   Vertical component of velocity = 61.41 m/s      

4 0
3 years ago
From the lens equation calculate the position of the following images produced by a convex lens.
fiasKO [112]

Explanation:

(i)

O is the object and I is the image.

The image formed is enlarged and it is erect. So the magnification will be positive (+) and greater than 1.

Refer above image. 1

(ii)

O is the object and I is the image.

The image formed is diminished and erect. So the magnification will be positive (+) and less than1.

Refer above image. 2

(iii)

The image will be formed as the 2F on the other side of the lens and it will be of same of the object.

5 0
3 years ago
sound is a longitudinal wave, and its speed depends on the medium through which it propagates. in air, sound travels at 343 m/s
melomori [17]

The speed of the sound wave in the medium, given the data is 3900 m

<h3>Velocity of a wave </h3>

The velocity of a wave is related to its frequency and wavelength according to the following equation:

Velocity (v) = wavelength (λ) × frequency (f)

v = λf

With the above formula, we can obtain the speed of the sound wave. Details below:

<h3>How to determine speed of the sound wave</h3>

The speed of the wave can be obtained as illustrated below:

  • Frequency (f) = 600 Hz
  • Wavelength (λ) = 6.5 m
  • Velocity (v) =?

v = λf

v = 6.5 × 600

v = 3900 m

Thus, the speed of the sound wave in the medium is 3900 m

Learn more about wave:

brainly.com/question/14630790

#SPJ4

8 0
1 year ago
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