Which terms both represent scalar quantities?

Calcula el peso aparente de una bola de aluminio de 50 cm3, cuando se encuentra totalmente sumergida en alcohol. Datos: la densi

artcher [175]

Answer:

W_apparent = 93.1 kg

Explanation:

The apparent weight of a body is the weight due to the gravitational attraction minus the thrust due to the fluid where it will be found.

W_apparent = W - B

The push is given by the expression of Archimeas

B = ρ_fluide g V

ρ_al = m / V

m = ρ_al V

we substitute

W_apparent = ρ_al V g - ρ_fluide g V

W_apparent = g V (ρ_al - ρ_fluide)

we calculate

W_apparent = 980 50 (2.7 - 0.8)

W_apparent = 93100 g

W_apparent = 93.1 kg

7 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:

1)

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):

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

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:

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

3)

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 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

( Image is 6 C carbon with the numbers 12.011 under it ) According to the image, the atomic mass of carbon is 12.011. How is the

Oduvanchick [21]

Answer:

B. By adding the number of protons and the number of neutrons

Explanation:

The atomic mass is determined by adding the number of protons and neutrons in an atom. An atom is made up of three fundamental particles: Electrons, Protons and Neutrons.

The protons and neutrons occupy a central region in an atom known as the nucleus. The nucleus is positively charged and mass concentrated.

If we compare the relative masses of the subatomic particles, the masses of protons and neutrons would be 1 and that of an electron would be 1/1840. This shows that the mass of electrons are negligible.

In order to ascertain atomic mass, we therefore add the number of protons and neutrons together. This is how we arrive at 12.011 as the value of the atomic mass of C and for other elements.

The atomic mass is also known as the mass number.

4 0

3 years ago

Imagine that two balls, a basketball and a much larger exercise ball, are dropped from a parking garage. If both the mass and ra

pashok25 [27]

Here if we assume that there is no air friction on both balls then we can say

$F = mg$