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3 years ago

Which of the following is an extensive property? Select one: a. Density b. Temperature c. Mass d. Specific Heat e. Pressure

Physics

1 answer:

weeeeeb [17]3 years ago

6 0

Answer:

C. Mass

Explanation:

The extensive property is that property of material that depends upon the amount of matter (mass) of substance. Thus, in the given options only, mass is the extensive property. All the other properties other than mass are independent of mass.

The simple method to find out the extensive property is to divide the substance into half. If that property for both individual halves also becomes half, then it is dependent on mass and an extensive property. If the property remains same as before for each half, then it is independent of mass and called intensive property such as, density, temperature, pressure, etc.

Thus the correct answer is <u>C. mass</u>

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nataly862011 [7]

Lực hút trái đất gọi Là trọng lực

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3 years ago

An 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching th

ddd [48]

Answer:

1626.4 N

Explanation:

Given that a 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What force does the water exert on him?

The parameters to be considered are:

Distance S = 3m

Time t = 0.55s

Since the man started from rest, initial velocity u = 0

Using second equation of motion

S = Ut + 1/2at^2

3 = 1/2 × a × 0.55^2

3 = 1/2 × a × 0.3025

a = 3/ 0.15125

a = 19.83 m/s^2

Force = mass × acceleration

Force = 82 × 19.83

Force = 1626.4 N

Therefore, the force that water exerted on him is 1626.4 N

4 0

2 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )