Which best describes what is made of matter?

Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH = 1.06

Slav-nsk [51]

Answer:

$1.06085\times 10^{-10}\ m$

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

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

e = Charge of electron = $1.6\times 10^{-19}\ C$

n = 1 (ground state)

Angular momentum is given by

$L=mvr$

From Bohr's atomic model we have

$L=\dfrac{nh}{2\pi}$

$mvr=\dfrac{nh}{2\pi}\\\Rightarrow v=\dfrac{nh}{2\pi mr}$

The centripetal force will balance the electrostatic force

$\dfrac{ke^2}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow \dfrac{ke^2}{r}=mv^2\\\Rightarrow \dfrac{ke^2}{r}=m(\dfrac{nh}{2\pi mr})^2\\\Rightarrow r=\dfrac{n^2h^2}{4\pi^2mke^2}\\\Rightarrow r=\dfrac{1^2\times (6.626\times 10^{-34})^2}{4\pi^2 \times 9.11\times 10^{-31}\times 8.99\times 10^{9}\times (1.6\times 10^{-19})^2}\\\Rightarrow r=5.30426\times 10^{-11}\ m$

The diameter is $2\times 5.30426\times 10^{-11}=1.06085\times 10^{-10}\ m$

7 0

3 years ago

A small 20-kg canoe is floating downriver at a speed of 2 m/s. What is the canoe’s kinetic energy? A. 40 J B. 80 J C. 18 J

rusak2 [61]

A small 20-kg canoe is floating downriver at a speed of 2 m/s. 40 J is the canoe’s kinetic energy.

Answer: Option A

<u>Explanation:</u>

The given canoe has the mass and is being given to move at a speed. Therefore the kinetic energy of the canoe can be calculated using the following method,

Given that mass of the canoe = 20 kg and its speed =1 m/s

As we know that the Kinetic energy has the formula,

$\text {Kinetic energy}=\frac{1}{2} \boldsymbol{m} \boldsymbol{v}^{2}$

Therefore, substituting the value into the equation, we get,

$K . E .=\frac{1}{2} \times 20 \times 2^{2}$ = 40 J

4 0

3 years ago

Read 2 more answers

Select the false statement. A. Only materials that have never been part of a living thing can be recycled in nature. B. Matter f

astra-53 [7]

Option A is the false statement.

Only materials that have never been part of a living thing can be recycled in nature.

In actual all living and non living thing can be recycled in nature.

3 0

3 years ago

The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70

MrRa [10]

Answer:

a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

C = $\epsilon_o \frac{A}{d}$

C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

C = $\frac{Q}{\Delta V}$

Q₀ = C ΔV

Q₀ = 2.62 10⁻¹² 8.70

Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

C₁ = 1.04 10⁻¹² F

C₁ = Q₀ / ΔV₁

ΔV₁ = Q₀ / C₁

ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

ΔV₁ = 21.9 V

b) initial stored energy

U₀ = $\frac{Q_o}{ 2C}$

U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)

U₀ = 99.2 10⁻¹² J

c) final stored energy

U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)

U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

W = U_f - U₀

W = (249.2 - 99.2) 10⁻¹²

W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0

2 years ago

A very weak pressure wave, i.e., a sound wave, across which the pressure rise is 30 Pa moves through air which has a temperature

Fofino [41]

Answer:

Density change, Δρ = 2.4 × 10⁻⁴ kg/m³

Temperature Change, ΔT = 0.0258 K

Velocity Change, Δc = 0.0148 m/s

Explanation:

For sound waves moving through the air,

Pressure and Temperature varies thus

(P₀/P) = (T₀/T)^(k/(k-1))

Where P₀ = initial pressure of air = 101KPa = 101000 Pa

P = final pressure of air due to the change brought about by the moving sound wave = 101000+30 = 101030 Pa

T₀ = initial temperature of air = 30°C = 303.15 K

T = final temperature of air = ?

k = ratio of specific heats = Cp/Cv = 1.4

(101000/101030) = (303.15/T)^(1.4/(1.4-1))

0.9990703 =(303.15/T)^(3.5)

Solving This,

T = 303.1758 K

ΔT = T - T₀ = 303.1758 - 303.15 = 0.0258 K

Density can be calculate in two ways,

First method

Δρ = ρ - ρ₀

P₀ = ρ₀RT₀

ρ₀ = P₀/RT₀

R = gas constant for air = 287 J/kg.k

where all of these are values for air before the wave propagates

P₀ = 101000 Pa, R = 287 J/kg.K, T₀ = 303.15K

ρ₀ = 101000/(287 × 303.15) = 1.1608655 kg/m³

ρ = P/RT

P = 101030 Pa, T = 303.1758K

ρ = 101030/(287×303.1758) = 1.1611115 kg/m³

Δρ = ρ - ρ₀ = 1.1611115 - 1.1608655 = 0.00024 kg/m³ = 2.4 × 10⁻⁴ kg/m³

Second method

(ρ₀/ρ) = (T₀/T)^(1/(k-1))

Where ρ₀ is initially calculated from ρ₀ = P₀/RT₀, then ρ is then computed and the diff taken.

Velocity Change

c₀ = √(kRT₀) = √(1.4 × 287 × 303.15) = 349.00669 m/s

c = √(kRT) = √(1.4 × 287 × 303.1758) = 349.0215415 m/s

Δc = c₀ - c = 349.0215415 - 349.00669 = 0.0148 m/s

QED!

5 0

2 years ago