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Umnica [9.8K]
3 years ago
12

The freezing point of water is the same as its a. melting point. b. boiling point. c. sublimation point. d. evaporation point.

Physics
2 answers:
MAXImum [283]3 years ago
8 0

Answer: melting point.

Explanation:

Freezing point is the temperature at which liquid phase changes to solid phase. Freezing point of water is 0^0C.

H_2O(l)\rightarrow H_2O(s)

Melting point is the temperature at which solid phase changes to liquid phase. Melting point of water is 0^0C.

H_2O(s)\rightarrow H_2O(l)

Melting and freezing are reversible processes which takes place at 0^0C for water.

Boiling point is the temperature at which vapor pressure of the liquid becomes equal to atmospheric pressure.

Sublimation point is the temperature at which gaseous phase changes to solid phase.

Thus the freezing point of water is the same as its melting point.

algol [13]3 years ago
6 0
A. Melting point .. I learned this today in class. 
Ex. Ice melts at 0 degrees C and freezes at 0 degrees C.
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Dehydration? I think that’s it.
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3 years ago
Two adjacent natural frequencies of an organ pipe are found to be 550. Hz and 650. Hz. (a) Calculate the fundamental fre- quency
liubo4ka [24]

Answer:

a) 100Hz

b) opened at both ends

c) 1.72m

Explanation:

Let fun be the nth Harmonic frequency organ. If pipe is opened at both ends.

L= length of pipe

Fyn =(n/x)

550hz= (n/2L)v 1

650hz=(n+1/2L)×v 2

Comparing 1 and 2

650hz= nv/2L + v/2L

550= 343/2L

L= 550×2/343 = 1.7m

c) fundamental frequency

Fo= nv/2L = 343/(2×1.7) = 100Hz

8 0
3 years ago
A wire is formed into a circle having a diameter of 10.0cm and is placed in a uniform magnetic field of 3.00mT . The wire carrie
Paul [167]

The range of potential energies of the wire-field system for different orientations of the circle are -

θ                  U

0°             375 π x 10^{-7}

90°              0

180°        - 375 π x 10^{-7}

We have current carrying wire in a form of a circle placed in a uniform magnetic field.

We have to the range of potential energies of the wire-field system for different orientations of the circle.

<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>

The formula to calculate the magnetic potential energy is -

U = M.B = MB cos $\theta

where -

M is the Dipole Moment.

B is the Magnetic Field Intensity.

According to the question, we have -

U = M.B = MB cos $\theta

We can write M = IA (I is current and A is cross sectional Area)

U = IAB cos $\theta

U = Iπr^{2}B cos $\theta

For $\theta = 0° →

U(Max) = MB cos(0) = MB =  Iπr^{2}B = 5 × π × ( 0.05 ) ^{2} × 3 × 10^{-3} =

375 π x 10^{-7}.

For $\theta = 90° →

U = MB cos (90) = 0

For $\theta = 180° →

U(Min) = MB cos(0) = - MB =  - Iπr^{2}B = - 5 × π × ( 0.05 ) ^{2} × 3 × 10^{-3} =

- 375 π x 10^{-7}.

Hence, the range of potential energies of the wire-field system for different orientations of the circle are -

θ                  U

0°             375 π x 10^{-7}

90°              0

180°        - 375 π x 10^{-7}

To solve more questions on Magnetic potential energy, visit the link below-

brainly.com/question/13708277

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3 0
1 year ago
A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the
Ierofanga [76]

Answer:

The discharge rate is Q = 0.0192 \  m^3 /s

Explanation:

From the question we are told that

   The  diameter is  d =  60 \ mm   =  0.06 \ m

    The  head is  h  =  6 \ m

     The  coefficient of contraction is  Cc  =  0.68

     The  coefficient of  velocity is  Cv  =  0.92

The radius is mathematically evaluated as

         r =  \frac{d}{2}

substituting values

        r =  \frac{ 0.06 }{2}

        r =  0.03 \ m

The  area is mathematically represented as

      A =  \pi r^2

substituting values

      A =  3.142 *  (0.03)^2

      A = 0.00283 \ m^2

 The  discharge rate is mathematically represented as

        Q =  Cv *Cc  *  A  *  \sqrt{ 2 * g *  h}

substituting values

       Q = 0.68 *  0.92*   0.00283  *  \sqrt{ 2 * 9.8 *  6}

       Q = 0.0192 \  m^3 /s

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3 years ago
a metal sphere with a mass of 90 kg rolls along at 16m/s and strikes a stationary sphere having a mass of 140kg. the first spher
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