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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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Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
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Answer: 3.7 \times 10^{-4} N

Explanation:

The gravitational pull between two object is given by:

F = G\frac{Mm}{r^2}

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N

4 0
3 years ago
Read 2 more answers
How much heat is needed to change the temperature of 3 grams of gold (c = 0.129 ) from 21°C to 363°C? The answer is expressed to
Temka [501]

Q= mcΔT

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You have to convert Celsius into kelvin in order to use this formula I believe

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Now...

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3 0
3 years ago
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A 40 kg person sits on top of a 400 kg rock. What is the person’s weight? 390 N
algol13

The weight of the person is given by:

W = mg

W = weight, m = mass, g = gravitational acceleration

Given values:

m = 40kg, g = 9.81m/s²

Plug in and solve for W:

W = 40(9.81)

W = 390N

7 0
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