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elena-s [515]
3 years ago
8

How do solids, liquids, and gases change from one state to another? What makes water (H2O) a simple type of matter to work with

here on Earth?
Physics
1 answer:
bazaltina [42]3 years ago
8 0

Answer:

All change in their form in relation to amount of heat energy.

Explanation:

  • Due to the chemicals bonds that are within these forms of matter. The solids are compact and have strong bonds where the atomic molecules are attached to each other firmly.  
  • While in gases the bonds are least developed and molecules are able to move freely with the least friction. In water, the bonding is neither compact nor loose.  
  • All this is possible due to the addition of heat to molecules as they gain energy they start to move. Hence changes their forms, while the water molecules one hydrogen bond that is attached to nonbonding pairs of electrons. The water molecules form and reform continuously.
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Help! PROJECTILE PROBLEM: A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25
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Here, 

height at failure, h1 = 525 m, 
upward acceleration, a = 2.25 m/s^2, 
velocity = v m/s, 
<span>
SO, </span>
<span>
v^2 = 2*a*h = 2*2.25*525 = 2362.5 </span>
Now, acceleration, g = 9.8 m/s^2, 
<span>
SO, </span>
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heigt, h1 = v^2/2g = 2362.5 / 2*9.8 = 120.54 meters </span>
Hence, 
<span>
a) </span>
Total height = 525+120.54 = 645.54 meters 

b) 
<span>time, for h1, t = v/g = sqrt(2362.5)/9.8 = 4.96 sec 

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7 0
3 years ago
I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXT
ratelena [41]

Answer:

The resistance that will provide this potential drop is 388.89 ohms.

Explanation:

Given;

Voltage source, E = 12 V

Voltage rating of the lamp, V = 5 V

Current through the lamp, I = 18 mA

Extra voltage or potential drop, IR =  E- V  

                                                    IR = 12 V - 5 V = 7 V

The resistance that will provide this potential drop (7 V) is calculated as follows:

IR = V

R = \frac{V}{I} = \frac{7 \ V}{18 \times 10^{-3} A} \ = 388.89 \ ohms

Therefore, the resistance that will provide this potential drop is 388.89 ohms.

7 0
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