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Serggg [28]
2 years ago
11

On Ramesh’s13th birthday, his father invited all his friends and their relatives. It was a big party with lots of food and DJs.

Ramesh didn’t like the loud sound of DJs and asked his father to play it in a low volume so that their neighbours do not get much disturbed and people at the party can also enjoy the music. Ramesh’s father felt good for his wisdom and did as he said.
→Do you think when loud music is played at a party is acceptable to all the people living in, neighbourhood? Give a reason for your answer
→How can you control noise pollution at your end?
Physics
1 answer:
Lerok [7]2 years ago
3 0

Answer:

The first fact to understand which carries the most weight is we share this small world. We all have such similar characteristics that if you listed them all in one column and then list traits that make each person an absolute individual, one of the lists would be liken to an encyclopedia where as the other more akin to a doodle on a napkin in comparison. Now to the question.

Explanation:

We all know inherently the rules, so to say. yes its acceptable to do as people do normally. to go outside those bounds on purpose to be a nuisance intentionally is not. like 3am loud pounding music or downright disrespect is not acceptable. because when its done that way, since we all know the unspoken rules, its is being done on purpose to annoy.

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refrigerant 134a enters a compressor operating at steady state as saturated vapor at 0.12 MPa and exits at 1.2 MPa and 70 C at a
Afina-wow [57]

Answer:

the power input to the compressor is 7.19Kw

Explanation:

Hello!

To solve this problem follow the steps below.

1. We will call 1 the refrigerant state at the compressor inlet and 2 at the outlet.

2. We use thermodynamic tables to determine enthalpies in states 1 and 2.

(note: Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  )

h1[quality=1, P=0.12Mpa)=237KJ/Kg

h2(P=1.2Mpa, t=70C)=300.6KJ/kg

3. uses the first law of thermodynamics in the compressor that states that the energy that enters a system is the same that must come out

Q=heat=0.32kJ/s

W=power input to the compressor

m=mass flow=0.108kg/S

m(h1)+W=Q+m(h2)

solving for W

W=Q+m(h2-h1)

W=0.32+0.108(300.6-237)=7.19Kw

the power input to the compressor is 7.19Kw

7 0
2 years ago
A slender uniform rod 100.00 cm long is used as a meter stick. Two parallel axes that are perpendicular to the rod are considere
Nataliya [291]

Answer:

The correct answer is D    I_{30} /I_{50} =   1.5

Explanation:

In this exercise the moment of inertia equation should be used

    I = ∫ r² dm

In addition to the parallel axis theorem

    I = I_{cm} + M D²

Where  I_{cm} is the moment of the center of mass, M is the total mass of the body and D the distance from this point to the axis of interest

Let's apply these relationships to our problem, the center of mass of a uniform rod coincides with its geometric center, in this case the rod is 1 m long, so the center of mass is in

    L = 100.00 cm (1m / 100 cm) = 1.0000 m

     x_{cm} = 50 cm = 0.50 m

Let's calculate the moment of inertia for this point, suppose the rod is on the x-axis and use the concept of linear density

    λ = M / L = dm / dx

    dm =  λ dx

Let's replace in the moment of inertia equation

    I = ∫ x² ( λ dx)

We integrate

    I =  λ x³ / 3

We evaluate between the lower limits x = -L/2 to the upper limit x = L/2

    I =  λ/3 [(L/2)³ - (-L/2)³] = lam/3  [L³/8 - (-L³ / 8)]

    I =  λ/3  L³/4

    I = 1/12  λ L³

Let's replace the linear density with its value

    I = 1/12  (M/L)  L³

    I = 1/12  M L²

Let's calculate with the given values

   I = 1/12  M 1²

   I = 1/12 M

This point is the center of mass of the rod

    Icm = I = 1/12 M  = 8.333 10-2 M

Now let's use the parallel axis theorem to calculate the moment of resection of the new axis, which is 0.30 m from one end, in this case the distance is

    D = x_{cm} - x

    D = 0.50 - 0.30

    D = 0.20  m

Let's calculate

   I_{30} =I_{cm} + M D²

   I_{30} = 1/12 M + M 0.202

   I_{30} = M (1/12 + 0.04)

   I_{30} = M 0.123

To find the relationship between the two moments of inertia, divide the quantities

  I_{30} / I_{50} = M 0.123 / (M 8.3 10-2)

   I_{30} /I_{50} = 1.48

The correct answer is d 1.5

6 0
2 years ago
What mass of ice (in g) can be melted if 27.2 kJ of thermal energy are added at the freezing point? Use molar mass = 18.02 g/mol
san4es73 [151]

Answer : The mass of ice melted can be, 3.98 grams.

Explanation :

First we have to calculate the moles of ice.

Q=\frac{\Delta H}{n}

where,

Q = energy absorbed = 27.2 kJ

\Delta H = enthalpy of fusion of ice = 6.01 kJ/mol

n = moles = ?

Now put all the given values in the above expression, we get:

27.2kJ=\frac{6.01kJ/mol}{n}

n=0.221mol

Now we have to calculate the mass of ice.

\text{Mass of ice}=\text{Moles of ice}\times \text{Molar mass of ice}

Molar mass of ice = 18.02 g/mol

\text{Mass of ice}=0.221mol\times 18.02g/mol=3.98g

Thus, the mass of ice melted can be, 3.98 grams.

3 0
2 years ago
A system uses 2380 I of energy to do work as 12,900 j of heat are added to the system. The change in internal energy of the syst
sergey [27]
The internal energy of the system is characterized by the equation U = Q + W where U is the internal energy, Q is the heat and W is work. You are given 2,380 J of energy to do work as 12,900 J of heat so add up. The internal energy is 15,280J.



4 0
2 years ago
The force on a wire is a maximum of6.71 10-2 N when placed between the pole faces of a magnet.The current flows horizontally to
Taya2010 [7]

Answer:

B.   i=2.79A

C.   F=0.066N

Explanation:

A) By the right hand rule we have that

F=iL x B

F=iLBsin(α)

If the wire jump toward the observer the top pole face is the magnetic southpole.

B) The diameter of the pole face is 15cm. We can take this value as L (the length in which the wire perceives the magnetic field). Hence, we have

F=iLBsin(\alpha)\\\alpha=90°\\F=iLB\\i=\frac{F}{LB}=\frac{6.71*10^{-2}N}{(0.15m)(0.16T)}=2.79A

C) Now the length of the wire that feels B is

L=\frac{0.15m}{cos(10\°)}=0.152m

and the force will be (by taking the degrees between the magnetic field vector and current vector as 80°)

F=iLBsin(\alpha)\\F=(2.79A)(0.152m)(0.16T)(sin(80\°))=0.066N

I hope this is useful for you

regards

8 0
2 years ago
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