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2 years ago

5

A radioactive nucleus has a half-life of 5 x 108 years. Assuming that a sample of rock (say, in an asteroid) solidified right af

ter the solar system formed, approximately what fraction of the radioactive element should be left in the rock today?

1 answer:

Marina86 [1]2 years ago

5 0

Answer:

So the present mass will be 0.0017002 times of the initial mass of the sample.

Explanation:

Given:

half life of the given sample, $t'=5\times 10^8\ yrs$

We know that solar system came in to existence approximately $4.6\times 10^{9}$ years ago.

<u>Let the initial quantity on the formation of the solar system be, </u> $m_o$ <u />

Then the final quantity at present be:

$m=m_o\times (\frac{1}{2})^{(t/t')}$

$m=m_o\times (\frac{1}{2} )^{4.6\times 10^9\div(5\times 10^8)}$

$m=m_o\times 0.0017002$

So the present mass will be 0.0017002 times of the initial mass of the sample.

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Six dogs pull a two-person sled with a total mass of 220 kg. The coefficient of kinetic friction between the sled and the snow i

irina [24]

Answer:

a) 2457J

b) 558W

c) 337N

Explanation:

Assuming dogs started from rest.

$v=a*t\\t=\frac{a}{v}\\v=12\frac{km}{h}\frac{1000m}{km}*\frac{1h}{3600s}\\\\v=3.3m/s\\\\t=\frac{3.3m/s}{0.75m/s^2}\\t=4.4s$

and the displacement is given by:

$d=\frac{1}{2}*a*t^2\\d=7.3m$

Using the energy conservation formula:

$K_i+U_i+W_d+W_f=K_f+U_f$

Because the motion started from rest the initial kinetic energy is zero, the motion occurred in-ground level so the gravitational energy is zero too.

the work done by the friction force is given by:

$W_f=F_f*d*cos(\theta)\\W_f=\µ*m*g*d*cos(180)\\W_f=0.08*220kg*9.8m/s^2*7.3m*(-1)\\W_f=-1259J$

so:

$W_d=\frac{1}{2}*220kg*(3.3m/s)^2+1259J\\W_d=2457J$

The power is given by:

$P=\frac{W}{t}\\\\P=\frac{2457J}{4.4s}\\\\P=558W$

and the force exerted by the dogs:

$W_d=F_d*d*cos(\theta)\\F_d=\frac{W_d}{d*cos(0)}\\\\F=\frac{2457J}{7.3m*(1)}\\\\F=337N$

4 0

2 years ago

Difference between the fertility rate and mortality rate of developed and least development countries

dybincka [34]

Answer:

The replacement fertility rate is indeed only slightly above 2.0 births per woman for most industrialized countries (2.075 in the UK, for example), but ranges from 2.5 to 3.3 in developing countries because of higher mortality rates, especially child mortality. pls give brainlest

5 0

3 years ago

A wall clock has a minute hand with a length of 0.55 m and an hour hand with a length of 0.26 m. Take the center of the clock as

Lemur [1.5K]

Answer:

The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

Explanation:

Given that,

Length of minute hand = 0.55 m

Length of hour hand = 0.26 m

The time taken by the minute hand to complete one revelation is

$T= 3600\ sec$

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=\dfrac{2\pi}{T}$

Put the value into the formula

$\omega=\dfrac{2\pi}{3600}$

$\omega=0.001745\ rad/s$

We need to calculate the magnitude of the acceleration of the tip of the minute hand of the clock

Using formula of acceleration

$a=r\omega^2$

Put the value into the formula

$a=0.55\times(0.001745)^2$

$a=1.675\times10^{-6}\ m/s^2$

Hence, The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

3 0

2 years ago

An RL series circuit is connected to an ac generator with a maximum emf of 20 V. If the maximum potential difference across the

Rama09 [41]

If the maximum emf of the ac generator is 20 V and the maximum potential difference across the resistor is 16 V Then the maximum potential difference across the inductor is 4 V.

Calculation:

Step-1:

It is given that the RL circuit is connected to a 20 V ac generator. The maximum potential difference across the resistor is 16 V. It is required to find the maximum potential drop across the inductor.

Step-2:

The maximum emf of the generator is equal to the sum of the maximum potential difference across the resistor and the maximum potential difference across the inductor.

Therefore,

The maximum potential difference across the inductor + Maximum maximum potential difference across the resistor = Maximum emf of the generator

Thus,

Maximum maximum potential difference across the inductor + 16 V = 20 V

Therefore,

Maximum maximum potential difference across the inductor = 20 V - 16 V = 4 V

Learn more about potential differences across resistor and inductor here,

brainly.com/question/15715072

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5 0

1 year ago

A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

8 0

3 years ago