Which statement about electrons and atomic orbitals is not true?

12) A neutral atom has two, eight, eight electrons in its first, second, and third energy levels.This information ________.

pentagon [3]

Answer:

C.

Explanation:

Option C

The information given in the question tells us about the number of electrons in an atom and also the number of shells in the atom. So, we will come to know about the atomic number, size and chemical properties of the atom. But we cannot determine atomic mass. Atomic mass is a function of number of neutrons and protons.

3 0

3 years ago

A circular coil of wire having a diameter of 20.0 cm and 3000 turns is placed in the earth's magnetic field with the normal of t

Bess [88]

Explanation :

It is given that,

Diameter of the coil, d = 20 cm = 0.2 m

Radius of the coil, r = 0.1 m

Number of turns, N = 3000

Induced EMF, $\epsilon=1.5\ V$

Magnitude of Earth's field, $B=10^{-4}\ T$

We need to find the angular frequency with which it is rotated. The induced emf due to rotation is given by :

$\epsilon=NBA\omega$

$\omega=\dfrac{\epsilon}{NBA}$

$\omega=\dfrac{1.5}{3000\times 10^{-4}\times \pi (0.1)^2}$

$\omega=159.15\ rad/s$

So, the angular frequency with which the loop is rotated is 159.15 rad/s. Hence, this is the required solution.

3 0

3 years ago

1. A car is stopped at a red light. When the light turns green, the car starts

Lubov Fominskaja [6]

Answer:

Because of inertia (Newton's First Law of motion)

Explanation:

According to Newton's First Law of motion:

"An object at rest (or in motion at constant velocity) will tend to stay at rest (or tend to keep moving with same velocity) unless acted upon an unbalanced force"

In this problem, the object we are analyzing is the coffee cup.

At the beginning, the cup is at rest, together with the car.

Later, the car starts moving when the light turns green.

If we apply Newton's First Law of motion to the cup, we see that the coffee cup tends to keep its state of rest: for this reason, as the car moves forward, the coffee in the cup will spill backward, into the rear seat. This property of an object to mantain its state of motion is also called as inertia.

8 0

3 years ago

Three forces act on an object. Two of the forces are at an angle of 100◦to each other and have magnitude 25N and 12N. The third

seraphim [82]

Answer:

F₄ = 29.819 N

Explanation:

Given

F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N

F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N

F₃ = (0 i + 0 j + 4 k) N

Then we have

F₁ + F₂ + F₃ + F₄ = 0

⇒ F₄ = - (F₁ + F₂ + F₃)

⇒ F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N

The magnitude of the force will be

F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N

6 0

3 years ago

Air enters a turbine operating at steady state at 8 bar, 1600 K and expands to 0.8 bar. The turbine is well insulated, and kinet

kobusy [5.1K]

Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

$m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W$

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

$h_i = h_0 + WW = h_i -h_0$

Using the relation T-P we can find the final temperature:

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\$

$\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K$

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

$W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg$

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

4 0

3 years ago