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slamgirl [31]
3 years ago
5

How long will it take for a radioactive isotope with a decay constant of 0.15 (which means a half life of 4.6 days ) to decay to

5% of its original value?
Physics
1 answer:
jek_recluse [69]3 years ago
6 0
Radioactive decay is given by:
N = No x e^(-λt)
We know that N/No has to be 0.05
λ = 0.15
0.05 = e^(-0.15t)
t = ln(0.05)/(-0.15)
t = 19.97 days
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How long will it take you to travel 20 miles on a bus that drives 60 miles/h?
ValentinkaMS [17]

Answer:

<u>20 Minutes</u>

<u></u>

Explanation:

Well we know Mph (Miles per hour) is distance over time : \frac{distance}{time} \\

R (rate) = 60

d (distance) = 20

t (time) = Unknown

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

 R = \frac{d}{t}  

   ↓

60 = \frac{20}{t}

   ↓

 t = \frac{20}{60}

    ↓

 t = \frac{1}{3}  or 0.3333

<em>So basically it would take one third of an hour. Lets change these units to minutes.</em>

60 * 0.333333 = 20

<em>So it would take you </em><u><em>20 minutes</em></u><em> to drive 20 miles on a bus that drives 60 mph</em>

<em />

Hope that helps

<em>~Siascon~</em>

8 0
2 years ago
Can somebody tell me what time is it
kodGreya [7K]

Answer:

8:21am

Explanation:

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5 0
2 years ago
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A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 μC on each plate. The plates have a
butalik [34]

Answer:

The potential difference between the plates is 596.2 volts.

Explanation:

Given that,

Capacitance C=260\ pF

Charge q=0.155\ \mu\ C

Separation of plates = 0.313 mm

We need to calculate the potential difference between the plates

Using formula of potential difference

V= \dfrac{Q}{C}

Where, Q = charge

C = capacitance

Put the value into the formula

V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}

V=596.2\ volts

Hence,The potential difference between the plates is 596.2 volts.

7 0
3 years ago
Pls help on this one
lidiya [134]

Answer:

D

Explanation:

because 50.0/10.0 = 5.0

5 0
2 years ago
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A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
aivan3 [116]

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

And after 30 revolution

θ=30revolution

θ=30×2πrad

θ=60πrad

Final angular velocity is

ωf=18rev/sec

ωf=18×2πrad/sec

ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

(36π)²=(22π)²+2α×60π

(36π)²-(22π)²=120πα

Then, 120πα = 8014.119

α=8014.119/120π

α=21.26 rad/s²

Let. convert to revolution /sec²

α=21.26/2π

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b. Time Taken to complete 30revolution

θ=60πrad

∆θ= ½(wf+wi)•t

60π=½(36π+22π)t

60π×2=58πt

Then, t=120π/58π

t=2.07seconds

c. Time to reach 11rev/sec

wf=wo+αt

22π=0+21.26t

22π=21.26t

Then, t=22π/21.26

t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

5 0
3 years ago
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