(Not sure how many examples you need so I will put three for each)
Physical:
- As you now know, water in its natural condition is a colorless, odorless, and tasteless liquid. The hexagonal structure of water's crystals.
- The temperature at which a liquid's vapor pressure equals the pressure around it, turning the liquid into vapor, is known as the boiling point. We are aware that water reaches its boiling point at 100°C.
- The temperature at which a material transition from a liquid to a solid is known as the freezing point. The freezing point of water, which is 0°C or 32°F, is the temperature at which liquid water changes to solid ice.
Chemical:
- One of the most significant characteristics of water is its amphoteric tendency. Amphoteric refers to a substance's capacity to function as an acid or base. Water is neither acidic nor basic in its natural form. Its capacity to give and receive protons is the key justification. However, rainfall has a pH between 5.2 and 5.8, making it mildly acidic.
- Water is referred to be the all-purpose solvent. This is due to its chemical makeup, physical characteristics, high dielectric constant, and other factors that make it the most solvent material. It can attract other compound molecules, disabling their molecular forces and causing them to dissolve since hydrogen and oxygen both have positive and negative charges that are available.
- Water is a chemical molecule made up of two hydrogen atoms and one oxygen atom. The liquid condition of that substance is often referred to as water, and the solid and gas phases are respectively referred to as ice and steam.
Answer:
potential energy at origin is 
Explanation:
given data:
electric field E = 5*10^{6} N/C
at x = 43 cm, y = 28 cm
distance btween E and origin
potential energy per unit charge 
potential energy at origin is 2.57*10^{6} volt
<span>(a) -9.97 m/s
(b) x = 2.83
This is a simple problem in integral calculus. You've been given part of the 2nd derivative (acceleration), but not quite. You've been given the force instead. So let's setup a function for acceleration.
f''(x) = -8x N / 3.1 kg= -8x kg*m/s^2 / 3.1 kg = -2.580645161x m/s^2
So the acceleration of the body is now expressed as
f''(x) = -2.580645161x m/s^2
Let's calculate the anti-derivative from that.
f''(x) = -2.580645161x m/s^2
f'(x) = -1.290322581x^2 + C m/s
Now let's use the known velocity value at x = 2.0 to calculate C
f'(x) = -1.290322581x^2 + C
1
1 = -1.290322581*2^2 + C
11 = -1.290322581*4 + C
11 = -5.161290323 + C
16.161290323 = C
So the velocity function is
f'(x) = -1.290322581x^2 + 16.161290323
(a) The velocity at x = 4.5
f'(x) = -1.290322581x^2 + 16.161290323
f'(4.5) = -1.290322581*4.5^2 + 16.161290323
f'(4.5) = -1.290322581*20.25 + 16.161290323
f'(4.5) = -26.12903227 + 16.161290323
f'(4.5) = -9.967741942
So the velocity is -9.97 m/s
(b) we want a velocity of 5.8 m/s
5.8 = -1.290322581x^2 + 16.161290323
0 = -1.290322581x^2 + 10.36129032
1.290322581x^2 = 10.36129032
x^2 = 8.029999998
x = 2.833725463</span>
Answer:
Explanation:
Given
Work required to stretch 1 ft is 12 ft-lb
and we have to find work required to stretch 3 in.
i.e. 
divide (1)&(2)
