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Vinil7 [7]
3 years ago
15

Would tap water likely freeze and boil at those exact temperatures? (temperature freezes at 0 degree celius and boils at 100 deg

ree celus.
Physics
1 answer:
Kitty [74]3 years ago
6 0

No

Explanation:

Tap water will not boil and freeze at those exact temperature which are the boiling and freezing points.

O°C and 100°C are the boiling points of pure water.

Pure water is a pure substance with a definite composition made up of 2 atoms of hydrogen and 1 atom of oxygen.

Pure water boils at  100°C  and freezes at O°C ,

  • Tap water is not pure water.
  • It has some impurities in it which causes causes the elevation of the boiling point of liquids and widening of the boiling range.
  • Tap water is treated water that contains some other dissolved chemicals.

learn more:

Molecular formula brainly.com/question/4574324

#learnwithBrainly

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Why do we get dizzy when we spin?
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2 years ago
A weather balloon is inflated to a volume of 26.8 LL at a pressure of 744 mmHgmmHg and a temperature of 31.2 ∘C∘C. The balloon r
elena-s [515]

Answer:

43.96 L

Explanation:

We are given that

V_1=26.8 L

P_1=744mm Hg

T_1=31.2^{\circ} C=31.2+273=304.2K

P_2=385mmHg

T_2=-14.8^{\circ}=273-14.8=258.2K

We know that

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=\frac{P_1V_1T_2}{P_2T_1}

Substitute the values

V_2=\frac{744\times 26.8\times 258.2}{385\times 304.2}

V_2=43.96 L

Hence, the volume of balloon at -14.8 degree Celsius=43.96 L

5 0
3 years ago
Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it
erica [24]

<u>Answer:</u>

2N/cm

<u>Step-by-step explanation:</u>

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

F=kx

where, F is the force which is stretching or compressing the spring,

k is the spring constant; and

x is the distance the spring is stretched.

Substituting the given values to find the elastic constant  k to get:

F=kx

4=k(2)

k=\frac{4}{2}

k=2

Therefore, the elastic constant is 2 Newton/cm.

5 0
3 years ago
A bob attached to a string of length L = 1.25 m, initially found at the equilibrium
malfutka [58]

The maximum displacement angle of the bob is 13⁰.

The given parameters;

  • <em>Length of the pendulum, L = 1.25 m</em>
  • <em>Initial velocity of the bob, v = 0.8 m/s</em>

The maximum displacement of the bob is calculated by applying the principle of conservation of energy;

P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{0.8^2}{2\times 10} \\\\h = 0.032 \ m

The maximum displacement angle is calculated as follows;

cos \theta = \frac{L-h}{L} \\\\cos \theta = \frac{1.25 - 0.032}{1.25} \\\\\cos \theta = \frac{1.218}{1.25} \\\\cos \theta = 0.9744\\\\\theta = cos^{-1}(0.9744)\\\\\theta = 13\ ^0

Thus, the maximum displacement angle of the bob is 13⁰.

Learn more here:brainly.com/question/13981780

4 0
2 years ago
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