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Mumz [18]
3 years ago
10

A student is given a red and a blue liquid. The two samples of liquids are heated in identical beakers over identical Bunsen bur

ners. The red liquid shows a dramatically faster increase in temperature. Which of these is the most likely explanation for this?
Physics
2 answers:
Savatey [412]3 years ago
4 0
The red beaker has a smaller specific heat
Morgarella [4.7K]3 years ago
4 0

Explanation :

It is given that a student is given a red and a blue liquid. The two samples of liquids are heated in identical beakers over identical Bunsen burners.

It is observed that the red liquid shows a dramatically faster increase in temperature.

This observation can be explained in terms of specific heat. It is defined as the heat required to raise the temperature.

We know that the specific heat of any liquid is given by :

c=\dfrac{Q}{m\Delta T}

Where c is the specific heat.

m is the mass of the substance.

Q is the amount of heat.

\Delta T is the change in temperature.

So, it is clear that the specific heat capacity is inversely proportional to the temperature.

The better explanation for this can be the specific heat of the red liquid is less because it drastically increases in temperature.

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Answer:

the angle is about 67.79 degrees

Explanation:

We know that at its maximum height, the vertical component of the projectile's launching (initial) velocity (Vyi) is zero, so at that point it total velocity equals the horizontal component of the initial velocity (Vxi = 0.5 m/s)

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we can use this information to find the y component of the velocity at that height via the formula:

v_{yf}^2-v_{yi}^2=-2\,g\,\Delta y\\\\v_{yf}^2-v_{yi}^2=-2\,g\,(\frac{v_{yi}^2}{4\,g} )\\v_{yf}^2=v_{yi}^2-\frac{v_{yi}^2}{2} \\v_{yf}^2=\frac{v_{yi}^2}{2}

Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:

1=\sqrt{v_{yf}^2+0.5^2} \\1=\sqrt{\frac{v_{yi}^2}{2} +0.5^2}\\1^2=\frac{v_{yi}^2}{2} +0.5^2}\\1-0.5^2=\frac{v_{yi}^2}{2} \\2(1-0.5^2)=v_{yi}^2\\1.5 = v_{yi}^2\\v_{yi}=\sqrt{1.5} \\

Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:

tan(\theta)=\frac{v_{yi}}{v_{xi}} =\frac{\sqrt{1.5} }{0.5} \\\theta= arctan(\frac{\sqrt{1.5} }{0.5})=67.79^o

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