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4 years ago

14

State two uses of high specific heat of water

1 answer:

Nitella [24]4 years ago

7 0

It helps in regulating extremes in the environment.

It can be used as a car radiators.

<u>Explanation:</u>

The amount of heat needed to raise a certain amount of temperature to the water is known as the specific heat of the water. For one gram of water to increase 1 degree celsius, water needs to absorb 4.184 J of heat.

Some of the uses of high specific heat of water are as follows,

The temperature of the water will remain the same from day to night in ponds or sea. During day time, the hot air from the sea or pond rises while the cool air from the sea moves to take its place. During the night, the hot air from the land moves to take its place, giving rise to a land breeze and thus maintaining the environment.

Another example, When water is circulated throughout an engine, heat will get absorbed. This water is pumped to a radiator and then disposed to the surrounding air.

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The potential energy between two atoms in a particular molecule has the form U(x) = 2.1 x 8 − 5.2 x4 where the units of x are le

a_sh-v [17]

Answer:

$x\approx 0.948$

Explanation:

The correct formula for the potential energy between two atoms in a particular molecule is:

$U(x) = \frac{2.1}{x^{8}}-\frac{5.2}{x^{4}}$

Where $x$ is the distance.

According to the definitions of potential energy and work, as well as the Work-Energy Theorem and the Principle of Energy Conservation. The relation between that and related force is:

$F = -\frac{dU}{dx}$

The function is derived in terms of distance:

$F (x) = \frac{84}{5\cdot x^{9}} -\frac{104}{5\cdot x^{5}}$

Then, it is needed to find at least of x so that F(x) equals to 0.

$\frac{84}{5\cdot x^{9}}-\frac{104}{5\cdot x^{5}}=0$

$\frac{84}{x^{4}}-104 = 0$

$84-104\cdot x^{4} = 0$

$x=\sqrt[4]{\frac{84}{104} }$

$x\approx 0.948$

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3 years ago

What is the water table?

Setler [38]

The subsurface zone in which all openings of the rock are filled with water is called the zone of saturation.

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4 years ago

You have an unpolarized light source and you wish to send a beam of this light with intensity Io through a number of sheets of p

tekilochka [14]

Answer:

Explanation:

a ) The angle between the polarization axis of two adjecent sheet

= 90 / 3 = 30 degree.

The formula for intensity of polarised light from unpolarised light ( first transmission

I₁ = I₀ /2

I₀ is intensity of unpolarised light and I₁ is intensity of light after first time polarization .

The relation of I₁ and I₂ is as follows

I₂ = I₁ cos²30

= I₀/2 x3/4

=3 I₀/8

Relation between I₃ and I₂ is as follows

I₃ = I₂ cos²30

= 3I₀ / 8 x 3/4

= 9 I₀ / 32

= 0 .28 I₀

In case of stack of 4 plates

angle between two plates = 90/4 = 22.5 degree

I₁ = I₀ /2

I₂ = I₁ cos²22.5

= I₀ /2 x .85

I₃ = I₂ cos²22.5

= I₀ /2 x .85 x .85

= .36 I₀

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3 years ago

Calculate the recoil velocity in the horizontal direction, in meters per second, of a 1.25-kg plunger that directly interacts wi

mart [117]

Answer:

$v_1=-8.19\ m/s$ '

Explanation:

It is given that,

Mass of the plunger, $m_1=1.25\ kg$

Mass of the bullet, $m_2=0.0175\ kg$

Initially both plunger and the bullet are at rest, $u_1=u_2=0$

Final speed of the bullet, $v_2=585\ m/s$

Let $v_1$ is the final speed of the plunger. Using the conservation of momentum to find it. The equation is as follows :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

Since, $u_1=u_2=0$

$m_1v_1+m_2v_2=0$

$v_1=-\dfrac{m_2v_2}{m_1}$

$v_1=-\dfrac{0.0175\times 585}{1.25}$

$v_1=-8.19\ m/s$

So, the recoil velocity of the plunger is 8.19 m/s. Hence, this is the required solution.

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3 years ago

The red car drive on a motor way in 800 m/s.The

Black_prince [1.1K]

800/20=40
So the green car would be able to drive 40m

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3 years ago