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2 years ago

Physical and chemical properties of water

Physics

1 answer:

posledela2 years ago

6 0

(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.

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Lee pushes horizontally with a force of 75 n on a 36 kg mass for 10 m across a floor. calculate the amount of work lee did. answ

svlad2 [7]

To find work, you use the equation: W = Force X Distance X Cos (0 degrees)
Following the Law of Conservation of Energy, energy cannot be destroyed nor created.

So you would do 75 N x 10m x Cos (0 degrees)= 750 J

8 0

2 years ago

Read 2 more answers

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

3 years ago

List two diseases of the muscular system.

lesantik [10]

Answer:

muscular dystrophy and myasthenia gravis

4 0

3 years ago

Read 2 more answers

This will give brainliest

Troyanec [42]

Answer:

It is likely to be option B: 50 degrees and 50 degrees

Hope it helps.......... pls mark as brainliest.

6 0

2 years ago

You lift a 25-kg child 0.80 m, slowly carry him 10 m to the playroom, and finally set him back down 0.80 m onto the playroom flo

maksim [4K]

To solve this problem we will apply the work theorem which is expressed as the force applied to displace a body. Considering that body strength is equivalent to weight, we will make the following considerations

$\text{Mass of the child} = m = 25kg$

$\text{Acceleration due to gravity} = g = 9.81m/s^2$

$\text{Height lifted} = h = 0.80m (Upward)$

Work done to upward the object

$W = mgh$

$W = (25)(9.81)(0.8)$

$W = 196.2J$

Horizontal Force applied while carrying 10m,

$F = 0N$

$W = 0J$

Height descended in setting the child down

$h' = -0.8m (Downwoard)$

$W = mgh'$

$W = (25)(9.81)(-0.80)$

$W = -196.2J$

For full time, assuming that the total value of work is always expressed in terms of its symbol, it would be zero, since at first it performs the same work that is later complemented in a negative way.

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3 years ago

Physical and chemical properties of water​

Physical and chemical properties of water