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lions [1.4K]
1 year ago
7

Why does water require such a higher temperature to raise the pressure than neon?

Physics
1 answer:
kodGreya [7K]1 year ago
4 0

Water require such a higher temperature to raise the pressure than neon due to strong bonds between atoms of water.

<h3>Why does water require such a higher temperature to raise the pressure than neon?</h3>

Water require such a higher temperature to raise the pressure than neon because it has strong hydrogen bonds and having high specific capacity as compared to other atoms and molecules.

So we can conclude that water require such a higher temperature to raise the pressure than neon due to strong bonds between atoms of water.

Learn more about temperature here: brainly.com/question/24746268

#SPJ1

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What is an example of cultural change due to technology?
Nitella [24]
The correct answer to your question here is D
8 0
3 years ago
Calculate the location xcm of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Ea
Jet001 [13]

Answer:

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

Explanation:

Let suppose that planet and satellite can be treated as particles. The masses of Earth and Moon (m_{E}, m_{M}) are 5.972\times 10^{24}\,kg and 7.349\times 10^{22}\,kg, respectively. The distance between centers is 384,403 kilometers. The location of the center of mass can be found by using weighted averages:

\bar x = \frac{x_{E}\cdot m_{E}+x_{M}\cdot m_{M}}{m_{E}+m_{M}}

If x_{E} = 0\,km and x_{M} = 384,403\,km, then:

\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}

\bar x = 4.673\,km

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

8 0
2 years ago
Structures on a bird feather act like a diffraction grating having 7000 7000 lines per centimeter. What is the angle of the firs
Anestetic [448]

Answer:

θ = 28.9°

Explanation:

We are given;

Wavelength; λ = 602nm = 602 x 10^(-9) m

Lines per centimetre = 7000 /cm = 700000 /m

Thus, the distance between 2 adjacent lines is;

d = 1/700000 = 1.43 x 10^(-6) m

The angle at which diffracted light is formed is given by the formula

mλ = d sinθ

Where;

m is the mth order of the diffraction

λ is the wavelength of the incident light

d is the distance separating the centres of 2 adjacent slits

θ is the angle at which diffraction occurs

From the question, m is 1 because it says first order.

Thus, plugging in the relevant values into mλ = d sinθ, we have;

1 x 602 x 10^(-9) = 1.43 x 10^(-6) sinθ

sinθ = 602 x 10^(-9)/(1.43 x 10^(-6))

sinθ = 0.42098

θ = sin^(-1) 0.42098

θ = 28.9°

8 0
2 years ago
an 269 kg object is moved a distance of 1.9 m by a force if 580 j of work is done on the object what is the object acceleration
IrinaK [193]

Answer:

Explanation:

w=f*d=580*1.5=870J

7 0
2 years ago
A car moving with an initial speed v collides with a second stationary car that is one-half as massive. After the collision the
Mashutka [201]

Answer:

4v/3

Explanation:

Assume elastic collision by the law of momentum conservation:

m_1v = m_1v_1 + m_2v_2

where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively

Substitute m_2 = m_1/2 \& v_1 = v/3

m_1v = \frac{m_1v}{3} + \frac{m_1v_2}{2}

Divide both side by m_1, then multiply by 6 we have

6v = 2v + 3v_2

3v_2 = 4v

v_2 = \frac{4v}{3}

So the final speed of the second car is 4/3 of the first car original speed

5 0
3 years ago
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