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tangare [24]
3 years ago
13

When work is done by an applied force, the object's energy will change. In this Interactive, does the work cause a kinetic energ

y change or a potential energy change?
Physics
1 answer:
Alex73 [517]3 years ago
6 0

Answer:

Explanation:

According to the work energy theorem, the work done by all the forces by a body is equal to the change in kinetic energy of the body.

Work done = change in kinetic energy

W = Final kinetic energy - initial kinetic energy

So, work causes change in kinetic energy.

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What is the half-life of an isotope if after 30 days you have 31.25 g remaining from a 250 g beginning sample size?
Dima020 [189]

Answer:The time required for half of the original population of radioactive atoms to decay is called the half-life. The relationship between the half-life, T1/2, and the decay constant is given by T1/2 = 0.693/λ.

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2 years ago
Please help!!!!!!!!!!
wlad13 [49]

Answer:

C. Acceleration due to gravity

Explanation:

which is 9.8 m/s2 on Earth. The formula for calculating weight is F = m × 9.8 m/s2.

5 0
3 years ago
Suppose a small quantity of radon gas, which has a half-life of 3.8 days, is accidentally released into the air in a laboratory.
Daniel [21]

Answer:

1 day

Explanation:

Let the safe level = x

The current level = x + 0.2x = 1.2 x

Thus,

Half life = 3.8 days

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{3.8}\ days^{-1}

The rate constant, k = 0.1824 days⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

So,

\frac {[A_t]}{[A_0]} = x / 1.2 x = 0.8333

t = ?

\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.8333=e^{-0.1824\times t}

t ≅ 1 day

<u>Lab must be vacated in 1 day.</u>

4 0
3 years ago
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