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2 years ago

How far did Michael Phelps swim, if he swam at a speed of 10m/s in 30 seconds?

Physics

1 answer:

Setler [38]2 years ago

5 0

Answer: his total speed would be 3 seconds

Explanation: because if you mulimply 10 x 3 you get 30 meaning that your answer would be 3

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Lead atoms occupy a volume of 3 x 10-29 m3. Each atom contributes two free electrons. Calculate the Fermi velocity of lead.

Juliette [100K]

<u>Answer:</u> The Fermi velocity of lead is 64.4 km/s.

<u>Explanation:</u>

To calculate the Fermi velocity, we use the equation:

$V_f=\frac{h}{2\pi m_e}(\frac{3\pi^2N}{V})^{1/3}$

where,

h = Planck's constant = $6.62\times 10^{-34}Js$

$m_e$ = mass of electron = $9.1\times 10^{-31}kg$

N = Number of atoms present in per volume of atom multiplied by number of electrons present in given atom = $\frac{2\times N_A\times V}{M}$

$N_A$ = Avogadro's number = $6.022\times 10^{26}mol^{-1}$ (When the mass is in kilograms)

V = Volume = $3\times 10^{-29}m^3$

M = molecular weight of lead = 207.2 g/mol

Putting values in above equation, we get:

$V_f=\frac{6.62\times 10^{-34}}{2\times 3.14\times (9.1\times 10^{-31})}(\frac{3\times (3.14)^2\times (2\times 6.022\times 10^{23}\times 3\times 10^{-29})}{3\times 10^{-29}\times 207.2})^{1/3}$

$V_f=0.0644\times 10^6m/s=64.4km/s$ (Conversion factor: 1 km = 1000 m)

Hence, the Fermi velocity of lead is 64.4 km/s

4 0

3 years ago

The temperature of 2.0 g of helium is increased at constant volume by ΔT. What mass of oxygen can have its temperature increased

gavmur [86]

Answer:

m = 9.6 g

Explanation:

Thermal energy given to helium gas at constant volume is given as

$Q = nC_v \Delta T$

so here we have

$C_v = \frac{3}{2}R$

$n = moles$

$n = \frac{2}{4} = 0.5$

so we have

$Q = \frac{3}{2}R(0.5)\Delta T$

now we know that

for oxygen gas we have

$C_v = \frac{5}{2}R$

for same amount of heat we have

$Q = nC_v \Delta T'$

$\frac{3}{2}R(0.5)\Delta T = \frac{m}{32} (\frac{5R}{2}) \Delta T$

$m = \frac{0.75 \times 32}{2.5}$

$m = 9.6 g$

8 0

3 years ago

Which of the following changes will always increase the efficiency of a thermodynamic engine? Choose all correct statements.

Ilya [14]

Answer:B,C,D

Explanation:

Thermodynamic efficiency is given by

$\eta =1-\frac{T_C}{T-H}$

$\eta$ efficiency can be increased by Keeping $_c$ constant and increasing $T_H$

Keeping $T_H$ constant and decreasing $T_c$

by increasing $\Delta T=T_H-T_c$

by decreasing $\frac{T_C}{T_H}$ ratio

5 0

3 years ago

Read 2 more answers

A force of 1000 newtons was necessary to lift a rock. A total of 3000 joules of work was done. How far was the rock lifted?

CaHeK987 [17]

Answer:

Explanation:

3 meters

4 0

2 years ago

A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

$E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}$

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

$r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm$

The magnitude of the electric field is now given as;

$E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m$

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0

2 years ago