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sergeinik [125]
3 years ago
11

Properties of macroscopic systems! Tummarize your present understanding of the properties of macroscopic systems.

Physics
1 answer:
g100num [7]3 years ago
6 0

Answer:

Intensive and extensive properties.

Explanation:

The system properties which are measurable in the bulk is known as macroscopic properties.

Macroscopic system can be further classified in the two categories which are discussed below:

(a) Intensive properties: The properties which are mass independent. The examples of these are:- melting point, pressure, temperature, etc.

(b) Extensive properties: These property is strictly mass dependent or on quantity of matter. The examples of these are :- Volume, mole, etc.

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What kind of stars make up the halo?
Step2247 [10]
By definition, a halo is a part of a galaxy wherein it mainly consists of scattered stars forming bulges having a significantly spherical structure. In addition, the galactic structure is commonly made up of old and metal-rich stars forming clouds of gas and dirt at the spirals of a galaxy.
8 0
4 years ago
Read 2 more answers
Vector C has a magnitude of 24.6 m and points in the − y ‑ direction. Vectors A and B both have positive y ‑ components, and mak
frez [133]

Answer:

A= 61.35

B= -44.40

Explanation:

1. Using the components method we have:

A_{x}= A cos \alpha\\B_{x}= B cos \beta\\C_{x}= 0\\\\A_{y}= A sin \alpha\\B_{y}= B sin \beta\\C_{y}= 24.6\\

Considering that the vector sum A+B+C=0, then:

|V|=\sqrt{V_{x}^{2} +V_{y}^{2} }=0

Then:

V_{x} ^{2} =0; V_{x} =0\\V_{y} ^{2} =0; V_{y} =0

It means the value of x and y component is 0.

2. Determinate the equations that describe each component:

V_{x}= A cos \alpha -B cos \beta=0  (1)\\V_{y}= A sin \alpha +B sin \beta - C=0   (2)

Form Eq. (1):

A=B \frac{cos \beta}{cos \alpha}     (3)

Replacing A in Eq. (2):

(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-C=0\\(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-=C\\\\B(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)=C\\B=C(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)^{-1}     (4)

Replacing values of C, α and β in (4):

B= 24.6 (\frac{(cos 27.7)(sin 44.9)}{cos 44.9}+sin 27.7)^{-1}  \\B= -44.4

Replacing value of B in (3)

A=-44.40\frac{cos 27.7}{sin 49.9} \\A= 61.35

5 0
3 years ago
Please help me with the following question:
pishuonlain [190]

Answer: a. 667N

b. 665N

c. 54.5N

Explanation:

a) on the surface of the earth

W = mg

W = 68 × 9.81

= 667N

b) at the top of Everest (8848 m above sea level).

W =mg × R²/(R + H)²

W = 667 × [6378²/(6378 + 8.848)²

W = 665N

c) has 2 1/2 times the radius of the earth

W = mg × R²/(R + H)²

W = 667 × R²/(R + 2.5R)²

W = 54.5N

3 0
3 years ago
Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4
kati45 [8]

Answer:

velocity  = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = \frac{\pi }{4} × (4.50×10^{-2})²  × velocity

solve it we get

velocity  = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = \frac{\pi }{4} × 3 ×  (4.50×10^{-2})²  × velocity

velocity = 52.4 m/s

5 0
3 years ago
At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107
dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

K = - V\cdot \frac{dP}{dV}

Where:

K - Bulk module, measured in pascals.

V - Sample volume, measured in cubic meters.

P - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

-\frac{K \,dV}{V} = dP

This resultant expression is solved by definite integration and algebraic handling:

-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP

-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}

\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}

\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }

The final volume is predicted by:

V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }

If V_{o} = 1\,m^{3}, P_{o} - P_{f} = -10132500\,Pa and K = 2.3\times 10^{9}\,Pa, then:

V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }

V_{f} \approx 0.996\,m^{3}

Change in volume due to increasure on pressure is:

\Delta V = V_{o} - V_{f}

\Delta V = 1\,m^{3} - 0.996\,m^{3}

\Delta V = 0.004\,m^{3}

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0
3 years ago
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