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3 years ago

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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The innermost satellite of jupiter orbits the planet with a radius of 422 × 103 km and a period of 1.77 days. what is the mass o

FinnZ [79.3K]

The mass of Jupitar is obtained from the calculations as 5.8 * 10^-14 Kg.

<h3>What is the mass of Jupitar?</h3>

There are nine planets in the solar system and the sun lies at the enter of our solar system. This is the heliocentric model of the solar system.

Given that;

T^2 = GMr^3/4π

T = period

G = gravitational constant

r = radius

M = mass of Jupitar

Now;

1 day = 86400 seconds

1.77 days = 1.77 days * 86400 seconds/1 day

= 152928 seconds

Making M the subject of the formula;

M =4πT^2/Gr^3

M = 4 * 3.142 * (152928)^2/6.67 × 10^-11 * (422 × 10^9)^3

M = 2.9 * 10^11/5.0 * 10^24

M = 5.8 * 10^-14 Kg

Learn more about mass of a planet:brainly.com/question/13851553

#SPJ1

5 0

2 years ago

An infinitely long line of charge has a linear charge density of 7.50×10^−12 C/m . A proton is at distance 14.5 cm from the line

Nata [24]

Answer:

10.22 cm

Explanation:

linear charge density, λ = 7.5 x 10^-12 C/m

distance from line, r = 14.5 cm = 0.145 m

initial speed, u = 3000 m/s

final speed, v = 0 m/s

charge on proton, q = 1.6 x 10^-19 C

mass of proton, m = 1.67 x 10^-27 kg

Let the closest distance of proton is r'.

The potential due t a line charge at a distance r' is given by

$V=-2K\lambda ln\left (\frac{r'}{r} \right )$

where, K = 9 x 10^9 Nm^2/C^2

W = q V

By use of work energy theorem

Work = change in kinetic energy

$qV = 0.5m(u^{2}-v^{2})$

By substituting the values, we get

$V=\frac{mu^{2}}{2q}$

$-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}$

$- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }$

$- ln\left ( \frac{r'}{r} \right )=\frac{1.67 \times 10^{-27}\times 3000\times 3000}{4\times 9\times 10^{9}\times 1.6\times 10^{-19}\times 7.5\times 10^{-12} }$

$- ln\left ( \frac{r'}{r} \right )=0.35$

$\frac{r'}{r} =e^{-0.35}$

$\frac{r'}{r} =0.7047$

r' = 14.5 x 0.7047 = 10.22 cm

5 0

3 years ago

A projectile is fired horizontally from a gun that is 54.0 m above flat ground, emerging from the gun with a speed of 330 m/s. (

krok68 [10]

Explanation:

Given

height of building (h)=54 m

Projectile velocity=330 m/s

initial vertical velocity $(u_y)=0$

Initial horizontal velocity $(u_x)=330 m/s$

Time taken to cover vertical height of 54 m

$h=ut+\frac{gt^2}{2}$

$54=0+\frac{9.81\cdot t^2}{2}$

$t=\sqrt{\frac{54\times 2}{9.81}}$

t=3.31 s

Horizontal distance traveled in this time

$R_x=u_x\times t$

$R_x=330\times 3.31=1094.94 m$

Vertical component of velocity when it hits the ground

v=u+at

$v=0+9.81\times 3.31=32.47 m/s$

3 0

3 years ago

A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.

Airida [17]

Answer:

a) $v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )$ , b) $\Delta K = 9.559\times 10^{-5}\,J$

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

$(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v$