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Illusion [34]
3 years ago
13

An air conditioner operating at steady state maintains a dwelling at 70oF on a day when the outside temperature is 90oF. If the

rate of heat transfer into the dwelling through the walls and roof is 30,000 Btu/h, might a net power input to the air conditioner compressor of 3 hp be sufficient? If yes, determine the coefficient of performance. If no, determine the minimum theoretical power input, in hp.
Physics
1 answer:
torisob [31]3 years ago
8 0
30000 btuh /3413 btuh/kw. = 8.8 kw

8.8 kw/.746 kw/hp = 11.8 hp if COP is 1

11.8/3 hp (COP coefficient of performance) = 3.99 COP

>>>So yes a 3.0 hp compressor with a nominal COP of 4 will handle the 30,000 btuh load.

3.2 to 4.5 is expected COP range for an air cooled heat pump or a/c unit.
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avanturin [10]
The answer would be A. As the bat is swung, it gains kinetic energy. But once it hits the ball, it loses, or transfers, it’s kinetic energy to the ball.
5 0
3 years ago
If I have an object that starts from rest and reaches a velocity of 25m/s over a distance of 11m what is the acceleration of the
Elodia [21]

Answer:

Here is something that may help you!!

Explanation:

I found it in a cite (not that I'm plagiarizing, or anything).

3 0
3 years ago
Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0
3 years ago
The electrons in the beam of a television tube have an energy of 19.0 keV. The tube is oriented so that the electrons move horiz
igomit [66]

Answer:

The direction is due south

Explanation:

From the question we are told that

     The energy of the electron is E = 19.0keV = 19.0 *10^3 eV

      The earths magnetic field is B = 42.3 \muT = 42.3 *10^{-6} T

     

Generally the force on the electron is perpendicular to the velocity of the elecrton and the magnetic field and this is mathematically reresented as

          \= F = q (\= v * \=B)

On the first uploaded image is an  illustration of the movement of the electron

    Looking at the diagram  we can see that in terms of direction  the magnetic force  is

             \= F  =q(\=v * \= B)= -( -\r i * - \r k)

                = -(- (\r i * \r k))

generally  i cross k = -j

      so the equation above becomes

             \= F = -(-(- \r j))

                = - \r j

This show that the direction is towards the south  

 

3 0
3 years ago
Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms a
lawyer [7]

Answer:

This is because The energies of atoms are quantized.

Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed

5 0
3 years ago
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