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krek1111 [17]
3 years ago
5

Based on the ionic notation shown here, what has happened to this ion. ***List all that apply!!!*** Fe2+ *

Physics
1 answer:
elena55 [62]3 years ago
5 0

Answer:

atom has become cation

atom has lost 2 electrons

Explanation:

when atom loses electron they become positively charged and cation are positive ion

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Describe in terms of kinetic and potential energy what happens if an apple falls from a tree and comes to rest on the ground( wr
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Answer:

An apple hanging at a branch has potential energy due its position. It can be written as PE= mgh where m is the mass of the apple h is the distance between the apple and the ground and g is the acceleration due to gravity.

as the apple falls from the tree it loses its potential energy and gains kinetic energy due to the movement of the apple. Its kinetic energy will be given by KE= 1/2mv²  where m is the mass of the apple and v is the speed with which the apple falls.

As the apple falls the height or the distance reduces and PE becomes reduces. But it gains Kinetic energy due to its speed.

But when the apple falls to the ground and comes to rest its kinetic energy is converted to potential energy.

thus the total energy remains the same. it changes from one form to the other but remains unaltered.

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3 years ago
the mass of a lump of gold is constant everywhere but its weight isnt explain both in weight and mass
DiKsa [7]
That is very true, but what is the question asking you.
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3 years ago
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What are two benefits of scientists using a diagram to model the water cycle?
Sidana [21]

Explanation:

it can be used to show how the parts of the cycle relate to one another

8 0
2 years ago
2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

  • v is image distance
  • u is object distance, u is 10 cm
  • f is focal length, f is 5 cm

{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

  • v is image distance, v is 10 cm
  • f is focal length, f is 5 cm
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{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

  • Image is magnified
  • Image is erect or upright
  • Image is inverted
  • Image distance is identical to object distance.
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What is the acceleration for an object moving with constant speed? Explain your answer
Oduvanchick [21]

Answer:

What is the acceleration of an object moving at a constant speed?

The Meaning of Constant Acceleration

The data table above show an object changing its velocity by 10 m/s in each consecutive second. This is referred to as a constant acceleration since the velocity is changing by a constant amount each second.

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