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3 years ago

In the attachment there is a density column where there is colour

Physics

2 answers:

Ratling [72]3 years ago

7 0

The star is pink. Pink is the MOST dense on the list.

The red thing below the star is the foot of the tube that all the liquids and the star are in.

Anettt [7]3 years ago

4 0

Answer:

For your kind information, read the question thoroughly!

The bottom is the container part and not the liquid.

I hope it will be useful.

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For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that

Harrizon [31]

Answer:

The time taken is $t = 40007 sec$

Explanation:

From the question we are told that

The diameter of the egg is $d_e = 5.5cm = \frac{5.5}{100} = 5.5*10^{-2}m$

The initial temperature of egg the $T_e = 4.3^{o}C$

The temperature of the boiling water $T_b = 100^oC$

The heat transfer coefficient is $H = 800 W/m^2 \cdot K$

The final temperature is $T_e_f = 74^oC$

The thermal conductivity of water is $k = 0.607 W/m^oC$

The diffusivity of the egg $\alpha = 0.146 * 10^{-6} m^2 /s$

Using one term approximation

We have the

$\frac{T_e_f - T_b}{T_e - T_b} = Ae^{-\lambda ^2 \tau}$

The radius is $r = \frac{5.5*10^ {-2}}{2} =2.75*10^{-2}m$ Note that this radius is approximation to that of a real egg

Now we need to obtain the Biot number which help indicate the value of $A \ and \ \lambda$ to use in the above equation

The Biot number is mathematically represented as

$Bi = \frac{H r}{k}$

Substituting values

$Bi = \frac{800 * 2.75 *10^{-2}}{0.607}$

$= 36.24$

So for this value which greater than 0.1 the coefficient $\lambda_1 \ and \ A_1$ is

$\lambda = 3.06632$

$A = 1.9942$

Substituting this into equation 1 we have

$\frac{74- 100}{4.3 - 100} = 1.9942 e^{-(3.0632^2) \tau}$

$0.2717= 1.9942 e^{-(3.0632^2) \tau}$

$0.2717= 1.9942 e^{-9.383 \tau}$

$0.13624 = e^{-9.383 \tau}$

Taking natural log of both sides

$-1.993 = -9.383\ \tau$

$\tau = 0.2124$

The time required for the egg to be cooked is mathematically represented as

$t = \frac{\tau r^2}{\alpha }$

substituting value is

$= \frac{0.2124 * 2.75 *10^{-2}}{0.146 *10^{-6}}$

$t = 40007 sec$

8 0

3 years ago

Select whether the argument is an example of a deductive or an inductive argument:

viva [34]

The answer is a inductive

7 0

3 years ago

A gasoline engine receives 200 J of energy from combustion and loses 150 J as heat to the exhaust. What is its efficiency? 75% 2

choli [55]

50/200×100%=25% is answer the formula is usefull energy output divided by total energy provided into 100%

8 0

3 years ago

Read 2 more answers

Which of the following is not a characteristic of a creative person

Katarina [22]

<span>B.Extrinsic motivation </span>

7 0

3 years ago

Read 2 more answers

A radioactive material has a count rate of 400 per minute. It has a half life of 40 years. How long will it take to decay to a r

cestrela7 [59]

Answer:

160 years.

Explanation:

From the question given above, the following data were obtained:

Initial count rate (Cᵢ) = 400 count/min

Half-life (t½) = 40 years

Final count rate (Cբ) = 25 count/min

Time (t) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Initial count rate (Cᵢ) = 400 count/min

Final count rate (Cբ) = 25 count/min

Number of half-lives (n) =?

Cբ = 1/2ⁿ × Cᵢ

25 = 1/2ⁿ × 400

Cross multiply

25 × 2ⁿ = 400

Divide both side by 25

2ⁿ = 400/25

2ⁿ = 16

Express 16 in index form with 2 as the base

2ⁿ = 2⁴

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the time taken for the radioactive material to decay to the rate of 25 counts per minute. This can be obtained as follow:

Half-life (t½) = 40 years

Number of half-lives (n) = 4

Time (t) =?

n = t / t½

4 = t / 40

Cross multiply

t = 4 × 40

t = 160 years.

Thus, it will take 160 years for the radioactive material to decay to the rate of 25 counts per minute.

7 0

2 years ago