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3 years ago

8

The ________ is located inside the outdoor condenser unit and receives the low-pressure refrigerant vapor through the suction li

ne and squeezes it into a smaller volume at a higher pressure.

1 answer:

dusya [7]3 years ago

6 0

Answer:

Compressor

Explanation:

Compressor - the compressor is a unit that placed at the inner side of the condenser. The main function of the compressor is to increase the pressure by reducing the volume of air. And this reduced volume of air is compressed for the cleaning purpose of the air tube. While condenser is used for heat exchange which converts steam into liquid form.

It is advisable not to operate the compressor when the outside temperature is below 65 degrees F.

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How is the egg transferred from the ovary to the uterus? through muscle contractions by using special hairs by producing estroge

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The egg is transferred from the ovary to the uterus through muscle contractions called uterine contractions by using special hair like structures called cilia.

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what happens to light when it falls upon a material that has a natural frequency equal to the frequency of the light?

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The energy from the light is transferred to the material, causing it to vibrate and absorb the light.

What is energy?
In physics, energy is the quantitative quality that is transmitted to the a body or a physical system, and is discernible in the work performed as well as in the form of light and heat. The law of conservation states that although energy can change its form, it cannot be created or destroyed. Energy is indeed a conserved quantity. The International System of Units' (SI's) joule is the measurement unit for energy (J). A moving object's kinetic energy, a solid object's elastic energy, chemical energy caused by chemical reactions, and the potential energy that an object stores (for instance because of its position inside a field) are examples of common forms of energy.

When light falls upon a material that has a natural frequency equal to the frequency of the light, the light will be absorbed by the material. This is due to resonance, which occurs when the frequency of the light matches the natural frequency of the material. The energy from the light is transferred to the material, causing it to vibrate and absorb the light.

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1 year ago

A 1.00 -kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.62a). The object has

LekaFEV [45]

The distance D where the object comes to rest is 1.08.m.

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(c) the distance D where the object comes to rest.

$W_{total} =$ ΔKE ⇒ -0.25*1*9.8*D = 0-1/2*1* $2.3^{2}$

⇒ $D=\frac{0.5*2.3^{2} }{2.45}$

⇒1.08.m

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1 year ago

The curved line PQR is the velocity-time graph for a car starting from rest.

Elden [556K]

His. Curbs I b h bs. H b u b

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3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago