Introduction to Symbolic Answers, Part B, Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.

Thermal Conductors don't have to be hot to transfer heat, explain a situation when a ice cube would still transfer heat to anoth

barxatty [35]

An ice cube would transfer heat to another object whose temperature
is lower than zero°C (32°F).

A block of "dry ice" is sitting there at a temperature of -78°C (-109°F).
An ice cube helps to melt dry ice nice and fast.

If you could find a block of solid nitrogen, its temperature would be
63K (-210°C, -346°F). An ice cube would transfer heat to that baby
so fast that it would instantly boil.

6 0

3 years ago

3.5-m-diameter merry-go-round is rotating freely with an angular velocity of 0.78 rad/s . Its total moment of inertia is 1710 kg

Monica [59]

I dont get the question your asking

4 0

3 years ago

Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0

3 years ago

Solve the inequality 2(n+3) – 4<6. Then graph<br> the solution.

Aloiza [94]

The solution is 22 2(n+3)-4&6

6 0

3 years ago

Help !!!! Estimate the number of breaths taken by a person during 44 years?

Lisa [10]

44 x 12. I got the 12 from the total of 12 months in a year.

44 > 40

x
12 > 10
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The way my teacher taught me how to estimate is look at the neighbor to 44 and 12. The only time 44 can become 50, is when the neighbor is 5 or up. Same thing for 12. Now, multiply 40 and 10.

40 x 10 = 400.

Therefore, your estimate is 400.

The real answer is 520 breaths.

6 0

3 years ago