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3 years ago

Convert the speed of light 3.0x10^8 m/s to km/day

Physics

2 answers:

Aleksandr [31]3 years ago

7 0

Answer: $2.592 \times 10^{8}km/day$

$1 m = 0.001 km\\ 1 s= 1.157\times10^{-5} days\\ 1 m/s = \frac {0.001}{1.157\times10^{-5}} km/day = 86.4 km/day \\ 3.0\times 10^{8} m/s = 3.0\times 10^{8} m/s \times \frac {86.4 km/day}{1m/s} =2.592 \times 10^{8}km/day$

just olya [345]3 years ago

6 0

Answer:

$2592*10^7\frac{km}{day}$ .

Explanation:

We need to know that 1km = 1000 m= 10^3m and 1 day = 86400 = 864 * 10^2s.

$3*10^8 \frac{m}{s} = 3*10^8\frac{m}{s} \frac{86400 s}{1 day}$

$= 3*10^8*864*10^2\frac{m}{day} = 2592 * 10^{10} \frac{m}{day}$

Now, $2592 * 10^{10} \frac{m}{day}\frac{1km}{10^3m}$

$= 2592 * 10^{10}*10^{-3}\frac{km}{day} = 2592*10^7\frac{km}{day}$ .

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The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron

FrozenT [24]

Answer:

Speed of the electron is 2.46 x 10^8 m/s

Explanation:

momentum of the electron before relativistic effect = $M_{0} V$

where $M_{0}$ is the rest mass of the electron

V is the velocity of the electron.

under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = $M_{0}/ \sqrt{1 - \beta ^{2} }$

where

$\beta = V/c$

c is the speed of light = 3 x 10^8 m/s

V is the speed with which the electron travels.

The new momentum will therefore be

==> $M_{0}V/ \sqrt{1 - \beta ^{2} }$

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75 $M_{0} V$ = $M_{0}V/ \sqrt{1 - \beta ^{2} }$

the equation reduces to

1.75 = $1/ \sqrt{1 - \beta ^{2} }$

square both sides of the equation, we have

3.0625 = 1/ $(1 - \beta ^{2} )$

3.0625 - 3.0625 $\beta ^{2}$ = 1

2.0625 = 3.0625 $\beta ^{2}$

$\beta ^{2}$ = 0.67

β = 0.819

substitute for $\beta = V/c$

V/c = 0.819

V = c x 0.819

V = 3 x 10^8 x 0.819 = 2.46 x 10^8 m/s

6 0

3 years ago

N2 + O2 → 2NO N-N triple bond: 941 kJ/mol O-O double bond: 495 kJ/mol N-O bond: 201 kJ/mol

kiruha [24]

Answer:

$\large \boxed{\text{761 kJ}}$

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

N₂ + O₂ ⟶ 2NO

N≡N + O=O ⟶ 2O-N=O

Bonds: 2N≡N 1O=O 2N-O + 2N=O

D/kJ·mol⁻¹: 941 495 201 607

$\begin{array}{rcl}\Delta H & = & \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\& = & 2 \times 941 +1 \times 495 - (2 \times 201 + 2\times 607)\\&=& 2377 - 1616\\&=&\textbf{761 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{761 kJ}}$}.$