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valentinak56 [21]
3 years ago
11

PLEASE HELP use the image to answer the question. what is the name of the aluminum ion?

Physics
2 answers:
solong [7]3 years ago
8 0

<em>Answer:</em>

<em>Aluminum 3+</em>

<em>Sorry if im not correct</em>.

labwork [276]3 years ago
7 0

Answer:

Aluminium(3+)

Explanation:

im big brains

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The correct answer would be C. Unaware because it is the opposite of leery
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Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the
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I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)

Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.

The 2.0-kg block feels

• the downward pull of its own weight, (2.0 kg) <em>g</em>

• the upward normal force of the surface, magnitude <em>n₁</em>

• kinetic friction, mag. <em>f₁</em> = 0.30<em>n₁</em>, pointing in the negative horizontal direction

• the contact force of the larger block, mag. <em>c₁</em>, also pointing in the negative horizontal direction

• the applied force, mag. <em>F</em>, pointing in the positive horizontal direction

Meanwhile the 3.0-kg block feels

• its own weight, (3.0 kg) <em>g</em>, pointing downward

• normal force, mag. <em>n₂</em>, pointing upward

• kinetic friction, mag. <em>f₂</em> = 0.30<em>n₂</em>, pointing in the negative horizontal direction

• contact force from the smaller block, mag. <em>c₂</em>, pointing in the <u>positive</u> horizontal direction (this is the force that is causing the larger block to move)

Notice the contact forces form an action-reaction pair, so that <em>c₁</em> = <em>c₂</em>, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:

• net vertical force:

<em>n₂</em> - (3.0 kg) <em>g</em> = 0   ==>   <em>n₂</em> = (3.0 kg) <em>g</em>   ==>   <em>f₂</em> = 0.30 (3.0 kg) <em>g</em>

• net horizontal force:

<em>c₂</em> - <em>f₂</em> = 0   ==>   <em>c₂</em> = 0.30 (3.0 kg) <em>g</em> ≈ 8.8 N

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3 years ago
Make a VIR chart.<br> All voltage, resistance, and current.
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3 years ago
A 70.0 kg sprinter starts a race with an acceleration of 2.00 m/s2. If the sprinter accelerates at that rate for 29 m, and then
andrey2020 [161]

Answer:

t = 12.82s

Explanation:

F = m×a

  = (70)×(2)

  = 140 N

during the acceleration, the sprinter cover d = 29 m with time:

d = 1/2×at

29 = 1/2×(2)×t^2

t^2 = 29s

t = 5.39s

and attains the velocity of:

v = a×t

  = 2×5.39

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Then,to cover the last x = 80 m with a speed of 10.77 m/s in time:

t = x/v

 = 80/10.77

 = 7.43s

Therefore, it will take the sprinter 7.43 + 5.39 = 12.82s

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3 years ago
A body is traveling at 5.0 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion s
loris [4]

To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

By definition kinetic energy is defined as

KE = \frac{1}{2} mv^2

Where,

m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

m*V_i = m_1V_1+m_2V_2

Here,

m = Total mass (8Kg at this case)

m_1=m_2 = Mass each part

V_i = Initial velocity

V_2 = Final velocity particle 2

V_1 = Final velocity particle 1

The initial kinetic energy would be given by,

KE_i=\frac{1}{2}mv^2

KE_i = \frac{1}{2}8*5^2

KE_i = 100J

In the end the energy increased 100J, that is,

KE_f = KE_i KE_{increased}

KE_f = 100+100 = 200J

By conservation of the moment then,

m*V_i = m_1V_1+m_2V_2

Replacing we have,

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V_1+V_2 = 10

V_2 = 10-V_1(1)

In the final point the cinematic energy of EACH particle would be given by

KE_f = \frac{1}{2}mv^2

KE_f = \frac{1}{2}4*(V_1^2+V_2^2)

200J=\frac{1}{2}4*(V_1^2+V_2^2)(2)

So we have a system of 2x2 equations

V_2 = 10-V_1

200J=\frac{1}{2}4*(V_1^2+V_2^2)

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: V_1 = 10m/s

Then replacing in (1) we have that

PART B: V_2 = 0m/s

8 0
3 years ago
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