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gulaghasi [49]
2 years ago
15

You just calibrated a constant volume gas thermometer. The pressure of the gas inside the thermometer is 286.0 kPa when the ther

mometer is in contact with boiling water. The boiling water is under regular atmospheric pressure. What is the temperature of the thermometer, when it reads a pressure of 465.0 kPa?
Physics
1 answer:
diamong [38]2 years ago
5 0

Answer:

T_{2} = 606.69 K

Explanation:

In that the gas thermometer is a constant volume, it is satisfied that:

\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} }  

How the boiling water is under regular atmospheric pressure, then

T_{1} = 373 .15 K

Thus

\frac{286000}{373.15} = \frac{465000}{T_{2} }

T_{2} = 606.69 K

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Which of the following is not a component of a normal experiment?
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D. Validity group

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A validity group is not necessarily a component of a normal experiment.

A normal experiment is made up of the dependent variables, independent variables, and most times the control group.

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Instruction
Simora [160]

According to Edge, the answers is

<h3>3100 V</h3>

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<h3>200v</h3>

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3 years ago
The curvature of the helix r​(t)equals(a cosine t )iplus(a sine t )jplusbt k​ (a,bgreater than or equals​0) is kappaequalsStartF
4vir4ik [10]

Answer:

\kappa = \frac{1}{2 b}

Explanation:

The equation for kappa ( κ) is

\kappa = \frac{a}{a^2 + b^2}

we can find the maximum of kappa for a given value of b using derivation.

As b is fixed, we can use kappa as a function of a

\kappa (a) = \frac{a}{a^2 + b^2}

Now, the conditions to find a maximum at a_0 are:

\frac{d \kappa(a)}{da} \left | _{a=a_0} = 0

\frac{d^2\kappa(a)}{da^2}  \left | _{a=a_0} < 0

Taking the first derivative:

\frac{d}{da} \kappa = \frac{d}{da}  (\frac{a}{a^2 + b^2})

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} \frac{d}{da}(a)+ a * \frac{d}{da}  (\frac{1}{a^2 + b^2} )

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 + a * (-1)  (\frac{1}{(a^2 + b^2)^2} ) \frac{d}{da}  (a^2+b^2)

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - a  (\frac{1}{(a^2 + b^2)^2} ) (2* a)

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 -  2 a^2  (\frac{1}{(a^2 + b^2)^2} )

\frac{d}{da} \kappa = \frac{a^2+b^2}{(a^2 + b^2)^2}  -  2 a^2  (\frac{1}{(a^2 + b^2)^2} )

\frac{d}{da} \kappa = \frac{1}{(a^2 + b^2)^2} (a^2+b^2 -  2 a^2)

\frac{d}{da} \kappa = \frac{b^2 -  a^2}{(a^2 + b^2)^2}

This clearly will be zero when

a^2 = b^2

as both are greater (or equal) than zero, this implies

a=b

The second derivative is

\frac{d^2}{da^2} \kappa = \frac{d}{da} (\frac{b^2 -  a^2}{(a^2 + b^2)^2} )

\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} \frac{d}{da} ( b^2 -  a^2 ) + (b^2 -  a^2) \frac{d}{da} ( \frac{1}{(a^2 + b^2)^2}  )

\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} ( -2  a ) + (b^2 -  a^2) (-2) ( \frac{1}{(a^2 + b^2)^3}  ) (2a)

\frac{d^2}{da^2} \kappa = \frac{-2  a}{(a^2 + b^2)^2} + (b^2 -  a^2) (-2) ( \frac{1}{(a^2 + b^2)^3}  ) (2a)

We dcan skip solving the equation noting that, if a=b, then

b^2 -  a^2 = 0

at this point, this give us only the first term

\frac{d^2}{da^2} \kappa = \frac{- 2  a}{(a^2 + a^2)^2}

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum

the value of kappa is

\kappa = \frac{b}{b^2 + b^2}

\kappa = \frac{b}{2* b^2}

\kappa = \frac{1}{2 b}

3 0
3 years ago
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Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

F = \frac{mv^{2} }{r}

Bqv = \frac{mv^{2} }{r}

or Eq = \frac{mv^{2} }{r}

Assume that you want a velocity selector that will allow particles of velocity v⃗  to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

6 0
3 years ago
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