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AVprozaik [17]
3 years ago
13

When checking for noncondensables inside a recovery cylinder why should the technician allow the temperature of the cylinder to

stabilize to room temperature before taking a pressure reading?
Physics
1 answer:
77julia77 [94]3 years ago
3 0
Before taking a pressure reading, it is necessary for the technician to first allow the temperature of the cylinder to stabilize to room temperature because a comparison with a temperature-pressure chart is only valid and true when both temperature and pressure of the refrigerant are stable. 
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When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the
KonstantinChe [14]

Answer:

The ratio is  KE : TM  =  0.75

Explanation:

from the question we are told that

  The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical  energy of the mass is mathematically represented as

        TM  =  \frac{1}{2}  *  k  *  A^2

Here  k is the spring constant  ,  A is the total displacement of the  the mass  from maximum  compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

        TM  =  KE  + PE

=>     KE = TM  - PE

Here the potential  energy of the mass is mathematically represented as

     PE   = \frac{1}{ 2}  *  k *  [ x ]^2

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is  

      x = \frac{A}{2}

So

     PE   = \frac{1}{ 2}  *  k *  [ \frac{A}{2}  ]^2

So

      KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{2}  *  k  *  [\frac{A}{2} ]^2

=>    KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{8}  *  k  *  A ^2

=>    KE =  0.375  *  k  *  A^2

So the ratio of  KE :  TM is  mathematically represented as

       \frac{KE}{TM} =  \frac{0.375  k A^2 }{0.5 k A^2}

=>    \frac{KE}{TM} = 0.75

3 0
3 years ago
The force of the added water produces a torque on the dam. In a simple model, if the torque due to the water were enough to caus
Fittoniya [83]

Answer:

The appropriate response is "\tau=\frac{1}{6} PgLh^3". A further explanation is described below.

Explanation:

The torque (\tau) produced by the force on the dam will be:

⇒  d \tau=XdF

On applying integration both sides, we get

⇒  \tau = \int_{0}^{a}x pgL(h-x)dx

⇒     = pgL\int_{0}^{h}(h-x)dx

⇒     =pgL[\frac{h^3}{2} -\frac{h^3}{3} ]

⇒     =\frac{1}{6} PgLh^3

8 0
2 years ago
One cereal bar has a mass of 37 g. What is the mass of 6 cereal bars? Is that more than or less than 1 kg? Explain your answer.
GuDViN [60]

The mass is 222g.  No, it is less than 1kg.  There are 1000 grams in a kilogram so it would be 0.222kg.  Hope this Helps :D

7 0
3 years ago
Which type of reaction is this?<br> CH3COOH + H20 2 H+ + CH3C00
kumpel [21]

Answer:

dissociation of acetic acid (like vinegar) in water

Explanation:

CH3COOH is basically vinegar (aka acetic acid)

dissociation of acetic acid in water

so when you put vinegar in water it makes 2 hydrogen ions (H+) and acetate ( CH3C00 ) a chemical that used in making film

8 0
2 years ago
A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc
Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a   (→)

F - Ffriction = m*a  ⇒  F - (μ*N) = F - (μ*m*g) = m*a   ⇒  a = (F - μ*m*g)/m

⇒    a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x    ⇒    vf = √(vi² + 2*a*x)

⇒   vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0
3 years ago
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