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maksim [4K]
3 years ago
9

Nuclear decay occurs according to first-order kinetics. Cobalt-60 decays with a rate constant of 0.131 years−1. After 5.00 years

, a sample has a mass of 0.302 g. What was the original mass of the sample?
Physics
1 answer:
alexandr402 [8]3 years ago
8 0

Answer : The original mass of the sample was 0.581 grams.

Explanation :

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 0.131\text{ years}^{-1}

t = time passed by the sample = 5.00 years

a = original or initial amount of the reactant  = ?

a - x = amount left after decay process = 0.302 g

Now put all the given values in above equation, we get

5.00\text{ years}=\frac{2.303}{0.131\text{ years}^{-1}}\log\frac{a}{0.302g}

a=0.581g

Therefore, the original mass of the sample was 0.581 grams.

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Define motion also justify that rest and motion are related terms​
Damm [24]

Answer;

Motion: A body is said to be in motion if it changes its position with respect to its surroundings.

Explanation:

Rest and motion are the relative terms because they depend on the observer's frame of reference. So if two different observers are not at rest with respect to each other, then they too get different results when they observe the motion or rest of a body .

one example for each. Rest: If a body does not change its position with respect to its surroundings, the body is said to be at rest. ... Motion: A body is said to be in motion if it changes its position with respect to its surroundings.

6 0
3 years ago
A car speeding down the highway honks its horn, which has a frequency 392 Hz, but a resting bystander hears the frequency 440 Hz
Natali [406]

Answer:

37.42 m/s

Explanation:

We know that apparent frequency, \bar f is given by

\bar f=f\frac {V}{V-V_s} where f is the given frequency in this case 392, V is the speed of sound in air which is given as 343 and V_s is the speed of car which is unknown, \bar f is given as 440 Hz

440=392\times \frac {343}{343-V_s}\\343-V_s=392\times \frac {343}{440}=305.5818182\\V_s=343-305.5818182=37.41818182\approx 37.42 m/s

8 0
3 years ago
ADP binds to platelets in order to intiate the activation process. Two binding sites were identified on platelets, one with a Kd
12345 [234]

Answer:B) The binding site with a kid of7.9uM is the low affinity site.

Explanation:

The rate at which macro molecules like protein bind together is called affinity. The higher the affinity site the more readily these macro molecules fuses together. 7.9uM is a low affinity value than 359uM.

5 0
4 years ago
A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb.
lana [24]

In this problem we have the electric field intensity E:

E = 6.5 × 10^4 newtons/coulomb

We have the magnitude of the load:

q = 6.4 × 10 ^{-19} coulombs

We also have the distance d that the load moved in a direction parallel to the field 1.2 × 10^{-2} meters.

We know that the electric potential energy (PE) is:

PE = qEd

So:

PE = (6.4 × 10^{-19})(6.5 × 10^4)(1.2 × 10^{-2})

PE = 5.0 x 10^{-16} joules

None of the options shown is correct.

6 0
3 years ago
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alukav5142 [94]
-17.555m/s

first I found the time it took for jacks stone to reach the bottom, using the formula vf = vi + at, vf and vi are final and initial velocities.

then i found the velocity at 6.6m using vf^2 = vi^2 + 2ad
and I found the time it took to get to 6.6m, so that I knew how long Jill waited to throw her stone, I used the formula d = t(vi+vf)/2, then i done total time - the time she waited, to get the time it took for there stones to hit the ground at the same time.

then to find the initial velocity of her throw I used the formula d = vit + (at^2)/2
4 0
3 years ago
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