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aleksklad [387]
2 years ago
15

Does water not have anomalous expansion?​

Physics
2 answers:
sleet_krkn [62]2 years ago
6 0

Explanation:

The anomalous expansion of water is an abnormal property of water whereby it expands instead of contracting when the temperature goes from 4o C to 0o C, and it becomes less dense. ... The density becomes less and less as it freezes because molecules of water normally form open crystal structures when in solid form.

CaHeK987 [17]2 years ago
5 0
No. Water does not have anomalous expansion
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A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the
insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}

Req =3.03Ω , R1 =12.18Ω

\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}

\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}

\frac{1}{R2}=0.2479

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

3 0
3 years ago
Monochromatic light of a given wavelength is incident on a metal surface. however, no photoelectrons are emitted. if electrons a
kifflom [539]
If the object, ends up with a positive charge, then it is missing electrons. if it is missing electrons, then it must have been removed form the object during the rubbing process.
5 0
2 years ago
In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21
Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

z = 20km

And the velocity can be written as,

V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V = 980.55m/s

From the properties of standard atmosphere at altitude z = 20km temperature is

T = 216.66K

k = 1.4

R = 287 J/kg

Velocity of sound at this altitude is

a = \sqrt{kRT}

a = \sqrt{(1.4)(287)(216.66)}

a = 295.049m/s

Then the Mach number

Ma = \frac{V}{a}

Ma = \frac{980.55}{296.049}

Ma = 3.312

So front stagnation temperature

T_0 = T(1+\frac{k-1}{2}Ma^2)

T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)

T_0 = 689.87K

Therefore the temperature at its front stagnation point is 689.87K

6 0
3 years ago
A container is filled with water and the pressure at the container bottom is P. If the container is instead filled with a liquid
den301095 [7]
Eat ahah is when s susuehs
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3 years ago
A deuteron, 21H, is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a n
Naddik [55]

Answer:

1917723.40119 m/s

Explanation:

m = Mass of deuteron = (1.637+1.675)\times 10^{-27}=3.312\times 10^{-27}

k = Boltzmann constant = 1.381\times 10^{-23}\ J/K

T = Temperature = 2.94\times 10^8\ K

RMS velocity is given by

V_r=\sqrt{\dfrac{3kT}{m}}\\\Rightarrow V_r=\sqrt{\dfrac{3\times 1.381\times 10^{-23}\times 2.94\times 10^8}{3.312\times 10^{-27}}}\\\Rightarrow V_r=1917723.40119\ m/s

The RMS velocity of the deutrons is 1917723.40119 m/s

8 0
3 years ago
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