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siniylev [52]
2 years ago
13

An investigator collects a sample of a radioactive isotope with an activity of 490,000 Bq.48 hours later, the activity is 110,00

0 Bq. Part A For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution What is the half-life of the sample?
Physics
1 answer:
Daniel [21]2 years ago
3 0

Answer:

The correct answer is "22.27 hours".

Explanation:

Given that:

Radioactive isotope activity,

= 490,000 Bq

Activity,

= 110,000 Bq

Time,

= 48 hours

As we know,

⇒ A = A_0 e^{- \lambda t}

or,

⇒ \frac{A}{A_0}=e^{-\lambda t}

By taking "ln", we get

⇒ ln \frac{A}{A_0}=- \lambda t

By substituting the values, we get

⇒ -ln \frac{110000}{490000} = -48 \lambda

⇒    -1.4939=-48 \lambda

                 \lambda = 0.031122

As,

⇒ \lambda = \frac{ln_2}{\frac{T}{2} }

then,

⇒ \frac{ln_2}{T_ \frac{1}{2} } =0.031122

⇒ T_\frac{1}{2}=\frac{ln_2}{0.031122}

         =22.27 \ hours  

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3 years ago
how much does a bookshelf weigh if the movers are pushing it at a speed of 10 m/s^2 by applying 100 N force
Delicious77 [7]

Answer:

10 kg

Explanation:

Assuming a frictionless surface, then force F=ma where F is the applied force, m is the mass and a is acceleration. Making m the subject of the formula then m=\frac {F}{a}

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7 0
3 years ago
The forces that hold different atoms or ions together are
denis23 [38]

Answer:

Chemical bonds

Explanation:

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Therefore answer is --

Chemical bonds

 

7 0
3 years ago
It takes Mars about ____ Earth years to orbit the Sun.
OlgaM077 [116]
<span>At this distance, and with an orbital speed of 24.077 km/s, Mars takes 686.971 Earth days, the equivalent of 1.88 Earth years, to complete a orbit around the Sun. This eccentricity is one of the most pronounced in the Solar System, with only Mercury having a greater one (0.205). 

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7 0
2 years ago
Read 2 more answers
A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a
Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

Z_{R} = R

R = resistance

Z_{L} = jωL

ω = voltage source angular frequency, L = inductance

Z_{C} = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

Z_{R} = 217Ω

Z_{L} = j(220)(0.875) = j192.5Ω

Z_{C} = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = Z_{R} + Z_{L} + Z_{C}

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

v_{R} = I×Z_{R}

v_{L} = I×Z_{L}

v_{C} = I×Z_{C}

Given values:

I = (0.0569∠1.147+220t)A

Z_{R} = 217Ω = (217∠0)Ω

Z_{L} = j192.5Ω = (192.5∠π/2)Ω

Z_{C} = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

v_{R} = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

v_{L} = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

v_{C} = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

v_{R} = (12.35∠1.147+220t)V

v_{L} = (10.95∠2.718+220t)V

v_{C} = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

v_{R} = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

v_{L} = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

v_{C} = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0
2 years ago
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