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egoroff_w [7]
2 years ago
7

The speed of sound in air is approximately 340 m/s. The speed of light in air is approximately 3 x 108 m/s. If 10 seconds elapse

s between seeing a lightning strike and hearing the thunder, how far away was the lightning strike?
Physics
1 answer:
olya-2409 [2.1K]2 years ago
3 0

Answer:

3400 m

Explanation:

Both lightning and thunder happen at the same time but one is faster than the other. The distance traveled by a sound can be calculated from its speed such that;

 speed = distance/time, hence, distance = speed x time.

<em>For a thunder with 340 m/s speed and 10 seconds away from lightning, the distance between the thunder and the lightning can be calculated as</em>;

distance = 340 m/s x 10 s = 3400 m

     

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3 years ago
Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3
cupoosta [38]

Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

J = - D \frac{\Delta c}{\Delta x}

Where \frac{\Delta c}{\Delta x} is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

D= D_0 e^{\frac{E_d}{RT}}

Where D_0 is the proportionality constant,

R is the gas constant, T the temperature and E_d is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}

Rearrange the equation to get the value of temperature,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

We have all the values in our equation.

\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}

\Delta x = 9.7*10^{-3}m

E_d = 82000J

D_0 = 6.5*10^{-7}m^2/s

J = 3.2*10^{-9}m^2/s

R= 8.31Jmol^{-1}K

Substituting,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}

T=1118.07K=844\°C

4 0
3 years ago
In a milkans apparatus,an oil drop of weight 2.0×10^-15kg accquires two surplus electrons. when a Potential difference of 620 vo
andriy [413]

Answer:

r = 9.92 mm

Explanation:

Given that,

Mass of oil drop, m=2\times 10^{-15}\ kg

It acquires 2 surplus electrons, q = +2e =3.2\times 10^{-19}\ C

Potential difference, V = 620 V

Thie potential difference is applied between the pair of horizontal metal plates the drop is in equilibrium.

We need to find the distance between the plates.

At equilibrium,

mg = qE

Since, E = V/r (r is distance between plates)

mg=\dfrac{qV}{r}\\\\r=\dfrac{qV}{mg}\\\\r=\dfrac{3.2\times 10^{-19}\times 620}{2\times 10^{-15}\times 10}\\\\=0.00992\ m\\\\=9.92\ mm

So, the distance between the plates is 9.92 mm.

6 0
2 years ago
1.3 kg of water, at an initial temperature of 25oC, is heated at a rate of 100 W in a well-insulated container. How much time (i
Leviafan [203]

Answer:

68.25 minutes.

Explanation:

Power = Energy/time or

Power = Heat/time

P = Q/t....................... Equation 1

Q = Pt ........................ Equation 2

Where Q = quantity of heat, P = power, t = time.

Also,

Q = cm(t₂-t₁) ................. Equation 3

Where c = specific heat capacity of water, m = mass of water, t₁ = initial temperature of water, t₂ = final temperature of water.

substitute equation 2 into equation 3

Pt = cm(t₂-t₁) ............... Equation 4

make t the subject of the equation

t = cm(t₂-t₁)/P................ Equation 5

Given: P = 100 W, m = 1.3 kg, t₁ = 25 °C, t₂ = 100 °C.

Constant: 4200 J/kg.K

Substitute into equation 5

t = 1.3(4200)(100-25)/100

t = 409500/100

t = 4095 seconds

t = (4095/60) minutes = 68.25 minutes.

Hence the time it will take the water to reach boiling point = 68.25 minutes

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A booklet is a small book—with fewer pages and smaller dimensions than a “real” book, and usually a paper cover.

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