When the outermost energy level of an atom is full it tends to be?

Mady has an infinite number of balls and empty boxes available to her. The empty boxes, each capable of holding four balls, are

True [87]

Answer:

The number of balls in the box at 2010th step is 6 balls

Explanation:

Since the boxes are arranged from left to right, and since the empty boxes are capable of holding 4 balls at once.

Therefore, her steps would follow this pattern ;

00001 - 1st Step

0002 - 2nd Step

0003 - 3rd Step

0004 -4th Step

0010 - 5th Step

0011 - 6th Step

0021 - 7th step

0031 - 8th step

0041 - 9th step

Looking at this pattern, it is obvious this digits are in base 5 because once we count to 4 in a unit, we move to the next at 1.

Thus,at Mady's 2010th step, we'll convert to base 5,thus 2010 in base 5 is calculated as;

2010 ÷ 5 = 402 r 0, so Base 5 is now: _ _ _ _ 0.

402 ÷ 5 = 80 r 2, so Base 5 is now: _ _ _ 2 0.

80 ÷ 5 = 16 r 0, so Base 5 is now:

_ _ 0 2 0.

16 ÷ 5 = 3 r 1, so Base 5 is now: _ 1 0 2 0.

3 ÷ 5 = 0 r 3, so Base 5 is now: 3 1 0 2 0.

So 2010 in base 10 is 31020 in base 5. Thus,the number of balls in the box at 2010th step is equal to 3 + 1 + 0 + 2 + 0 = 6balls

3 0

3 years ago

80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting te

vodka [1.7K]

Answer:

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

$m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C$

Therefore, the resulting temperature is 23.37 ⁰C

6 0

3 years ago

A tiny boat is floating in a metal bucket. If you build a fire under the bucket ,how will most of the heat be transferred from t

Dimas [21]

The molecules will heat up and move faster, some evaporating and turning to gas, the toy boat will heat up if made of conducting materials but otherwise unchanged. The water will also start to boil.

7 0

3 years ago

Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli

Masja [62]

Answer:

Not possible

Explanation:

$E_{cl}$ = longitudinal modulus of elasticity = 35 Gpa

$E_{ct}$ = transverse modulus of elasticity = 5.17 Gpa

$E_m$ = Epoxy modulus of elasticity = 3.4 Gpa

$V_{\rho l}$ = Volume fraction of fibre (longitudinal)

$V_{\rho t}$ = Volume fraction of fibre (transvers)

$E_f$ = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

$E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764$

Transverse modulus of elasticity is given by

$E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148$

$V_{\rho l}\neq V_{\rho t}$

Hence, it is not possible to produce a continuous and oriented aramid fiber.

5 0

3 years ago

Can you please help me find the answer

Serggg [28]

Im sorry may you please retake the picture then i will answer

8 0

3 years ago