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2 years ago

Physics question, any help is appreciated :)

Physics

2 answers:

Gekata [30.6K]2 years ago

8 0

Runner 2 sees Runner 1 passing him with a velocity of 17 m/s west.

Rainbow [258]2 years ago

4 0

The velocity of 17 m/s west is the answer.

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How can a heat pump warm a house by causing a refrigerant to evaporate and condense

sveta [45]

The hot discharge gas from the refrigerant compressor is normally cooled and condensed at high pressure. This is then passed through an 'Expansion' valve which decreases the pressure to a low level causing expansion of the refrigerant liquid.
The liquid partially vapourises causing a 'Joule's/Thompson' refrigeration effect' which decreases temperature of the refrigerant which then passes to an evaporator coil in the air circulation system of the building. 
In the evaporator coil, the heat exchange between the cold refrigerant and the warm air of the building, vaporises and heats the refrigerant which returns to the compressor. 
The cycle is repeated until the air temperature reaches the thermostat set-point and switches off the system. 

As a Heat pump, the hot refrigerant gas is not evaporating and condensing. 
From the compressor discharge, the hot gas is by-passing the cooler/condenser unit and the expansion valve and passes directly to the 'evaporator' coils but now, as the heating medium for the air circulation system where it's cooled by the heat exchange between the hot gas and the cooler air in the building and returns to the compressor in a continuous cycle. 

A Thermostat in the system starts and stops the compressor motor according to the heat or cool temperature settings.

4 0

2 years ago

Determine the ratio of the flow rate through capillary tubes A and B (that is, Qa/Qb).

Oliga [24]

To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.

It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.

The mathematical equation that expresses this concept is

$\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}$

Where

P = Pressure at each point

r = Radius

$\eta =$ Viscosity

l = Length

Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}$

From the problem two terms are given

$R_A = \frac{R_B}{2}$

$L_A = 2L_B$

Replacing we have to

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}$

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}$

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}$

Therefore the ratio of the flow rate through capillary tubes A and B is 1/32

6 0

2 years ago

When a potential difference of 154 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge d

tester [92]

Answer:

$4.7\mu m$

Explanation:

We are given that

Potential difference=V=154 V

Surface charge density= $\sigma=29nC/m^2=29\times 10^{-5} C/m^2$

Using $1 nC/cm^2=10^{-5} C/m^2$

We know that

$C=\frac{\epsilon_0A}{d}=\frac{Q}{V}=\frac{\sigma A}{V}$

$d=\frac{\epsilon_0V}{\sigma}$

Where

$\epsilon_0=8.85\times 10^{-12}C^2/Nm^2$

Using the formula

$d=\frac{8.85\times 10^{-12}\times 154}{29\times 10^{-5}}$

$d=4.7\times 10^{-6} m=4.7\mu m$

Using $1\mu m=10^{-6} m$

Hence, the spacing between the plates= $4.7\mu m$

3 0

2 years ago

How much extra water does a 147-lb concrete canoe displace compared to an ultralightweight 38-lb kevlar canoe of the same size c

romanna [79]

We use the formula, to calculate the volume of water displaced by concrete canoe,

$V =\frac{W}{\gamma }$

Here, W is the weight of concrete canoe and $\gamma$ is the specific weight of water and its value is $62.4\ lb/ft^3$ .

So,

$V =\frac{ 147\ lb}{62.4\ lb/ft^3}=2.356\ ft^3$ .

Now the volume of water occupied in ultra lightweight kevlar canoe,

$v=\frac{w}{\gamma}$

Here, w is weight of kevlar canoe.

So,

$v=\frac{38\ lb}{62.4\ lb/ft^3} =0.6089\ ft^3$

Thus, the volume of water displaced,

$=V-v=2.356\ ft^3-0.6089\ ft^3=2.19\ ft^3$ .

Hence, the volume of water displaced canoe compared to an ultra-lightweight kevlar canoe is $2.19\ ft^3$

5 0

2 years ago

Describe metallurgy. Check all that apply.

mylen [45]

Answer:

A, C, D

Explanation:

got it right

7 0

2 years ago