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3 years ago

12

A uniform solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with an angular speed of 33.4

rpm. What is its kinetic energy?

1 answer:

DENIUS [597]3 years ago

5 0

Answer:

Is the equation for Ec=1/2 m(Dv)^2 where Dv is the difference between the angular speed & the areolar speed?

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Calculate the amount of heat (in kJ) required to convert 97.6 g of water to steam at 100° C. (The molar heat of vaporization of

Ratling [72]

Answer:

221.17 kJ

Explanation: Note the heat of vaporization is in kJ/mol,then to determine the number of moles of water: divide the mass by 18. Then multiply the number of moles by the molar heat of vaporization of water.

N = 97.6 ÷ 18

Q=molar heat *moles

Q = (40.79) * (97.6 ÷ 18)

This is approximately 221.17 kJ

4 0

2 years ago

A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact

ArbitrLikvidat [17]

Answer:

0.09 % of the original radioactive nucllde its left after 10 half-lives
It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

$N ( t) \ = \ N_0 \ e^{ \ - \frac{t}{\tau}}$ ,

where N(t) its quantity of material at time t, $N_0$ its the initial quantity of material and $\tau$ its the mean lifetime of the radioactive element.

The half-life $t_{\frac{1}{2}}$ its the time at which the quantity of material its the half of the initial value, so, we can find:

$N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

so:

$\ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

$e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}$

$- \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )$

$t_{\frac{1}{2}}\ = \tau ln( 2 )$

So, after 10 half-lives, we got:

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ t_{\frac{1}{2}}}{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ \tau \ ln( 2 ) }{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - 10 \ \ ln( 2 ) }$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}$

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

$10 \ * \ 24,110 \ years \ = \ 241,100 \ years$

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

7 0

3 years ago

Read 2 more answers

1. What are the three ways an object can accelerate?

solong [7]

Answer:

increase speed, decrease speed, and change direction

Explanation:

4 0

2 years ago

A persons beliefs and general outlook, which Act like filters on the information they receive are called ?

lianna [129]

The correct answer is:

Consideration

<em>Consideration</em> is defined as the act of thinking something carefully, careful thought; meditation; deliberation. Consideration acts like filter on the information a person receives, because it allows them to contemplate the implications of something before acting on it.

7 0

3 years ago

The intensity of electromagnetic radiation from the sun reaching the earth's upper atmosphere is 1.37kW/m2kW/m2. Part A Assuming

Juli2301 [7.4K]

Answer:

#_photon = 7 10²¹ photons

Explanation:

Let's look for the power that affects the panel of area of 1.5 m2

I = P / A

P = I A

P = 1.37 10³ 1.5

P = 2,055 10³ W

P = E / t

If we use t = 1 s

E = P t

E = 2,055 10³ J

This is the power that the panel receives, let's look for the energy of a photon

E = h f

c = λ f

f = c /λ

E = h c /λ

Let's calculate

E₀ = 6.63 10⁻³⁴ 3 10⁸/680 10⁻⁹

E₀ = 2.925 10⁻¹⁹ J

In one second the total energy is the number of photons for the energy of each one

E = #_photon E₀

#_photon = E / E₀

#_photon = 2,055 10³ / 2,925 10⁻¹⁹

#_photon = 7 10²¹ photons

8 0

2 years ago