A car travels 90 km/h. How long does it take for the car to travel

3. According to Hund's rule, what's the expected magnetic behavior of vanadium (V)?

ivanzaharov [21]

Answer:

Diamagnetic

Explanation:

Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).

For vanadium V ion, there are 18 electrons which will be arranged as follows;

1s2 2s2 2p6 3s2 3p6.

All the electrons present are spin paired hence the ion is expected to be diamagnetic.

6 0

3 years ago

You throw a rock horizontally out of a 7th story window. You time that it takes 3.7 seconds to hit the ground, and measure that

Nady [450]

Since we are only looking at the vertical height, we can use the free fall equation to find the height:
h = 0.5*g*t^2, where h is height in m, g is acceleration due to gravity (9.81 m/s^2), and t is time in seconds
h = 0.5*(9.81 m/s^2)*(3.7 s)^2
h = 67.15 m
Therefore, the 7th floor window is 67.15 m above ground level.

5 0

3 years ago

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

8 0

3 years ago

Jupiter and Saturn each have the same three basic cloud layers, but the spacing of the layers differs on the two planets. Why is

-Dominant- [34]

Answer:

b. Jupiter’s greater gravity has compressed the layers, so they are closer together there.

Explanation:

The value for Jupiter mass is 1.8981×10²⁷kg, while the mass of Saturn is 5.6832×10²⁶kg, so the different layers of clouds in Jupiter will be submitted to a greater gravitational pull because it has a bigger mass, as is established in the law of universal gravitation:

$F = G\frac{m1m2}{r^{2}}$ (1)

Where m1 and m2 are the masses of two objects, G is the gravitational constant and r is the distance between the two objects.

As it can be seen in equation 1, the gravitational force is directly proportional to the product of the masses of the objects, so if the mass increase the gravitational force will do it too.

For the case of Saturn, it has a lower mass so its layers of clouds will suffer a weaker gravitational pull. That leads to the three clouds being more spacing that the ones of Jupiter.

3 0

3 years ago

Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep

SIZIF [17.4K]

Answer:

1) iii i= 1m, 2) iii and iv, 3) i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

f = r / 2

f = 2/2

f = 1 m

The builder's equation

1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

1 / f = 0 + 1 / i

i = f

i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

R₂ = infinity

R₁ > 0

Cas2 Flat-concave (divergent) lens

R₂ = infinity

R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

1 / f₂ = 1 / (L-f₁) + 1 / i

1 / i = 1 / f₂ - 1 / (L-f₁)

1 / i = (L-f₁-f₂) / f₂ (L-f₁)

i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

6 0

3 years ago