At a distance of one centimeter from an electron, the electric field strength has a value

1 answer:

6 0

We are asked in the problem to determine the distance of an electron in which the electric field strength is equal to e/3. we use the gravitational formula of E = m1m2/d^2
hence,

e = m1m2/1 = k/1 =k
e/2 = k/d^2
e/2=e/d^2
d^2 = 2
d = sqrt 2 = 1.41 cm

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A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi

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Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

$m_{m}$ = mass of metal = 400 g

$c_{m}$ = specific heat of metal = ?

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$c_{a}$ = specific heat of aluminum cup = 900.0 J/kg ∙ K

$T_{ai}$ = initial temperature of aluminum cup = 15 °C

$m_{w}$ = mass of water = 500 g

$c_{w}$ = specific heat of water = 4186 J/kg ∙ K

$T_{wi}$ = initial temperature of water = 15 °C

$T$ = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

$m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}$

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