Answer:
1694 days
Explanation:
In first-order kinetics, the rate is proportional to the amount.
dA/dt = kA
For first-order kinetics, the rate k can be found using the half-life:
t₁,₂ = (ln 2) / k
In other words, the half-life is inversely proportional with the rate.
At the lower temperature, the rate is reduced to a third, so the half-life increases by a factor of 3. Meaning that the new half-life is 170 × 3 = 510 days.
The "shelf life" is the time it takes to reduce the initial amount to 10%. We can solve for this using the half-life equation.
A = A₀ (½)^(t / t₁,₂)
A₀/10 = A₀ (½)^(t / 510)
1/10 = (½)^(t / 510)
ln(1/10) = (t / 510) ln(½)
ln(10) = (t / 510) ln(2)
ln(10) / ln(2) = t / 510
t = 510 ln(10) / ln(2)
t ≈ 1694
The one word you're looking for to fill in the blank
can be "uneven" or "non-uniform".
Answer:
Explanation:
radius of circle r = 0.9 m.
(a ) In a motion on circular path , work done is zero because force ( centripetal force ) acts perpendicular to displacement .
( b )
Tension in string T = m ω²r
Putting the values
60 = .072 x ω² x 0.9
ω² = 926
ω = 30.4 rad /s
angle made in 20 revolutions θ = 20 x 2π = 126.6 rad
time taken = θ / ω
= 126.6 / 30.4
= 4.16 s .
answer:
heating the material
placing the material in a magnetic field of opposite polarity
hitting the material
The car will take 300 m before it stops due to applying break.
<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
- As per Newton's equation of motion, V² - U² = 2aS
- V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
- Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
- So, 0² - 60² = 2×6× S
=> -3600 = -12S
=> S = 3600/12 = 300 m
Thus, we can conclude that the distance covered by the car is 300 m before it stopped.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?
Learn more about the Newton's equation of motion here:
brainly.com/question/8898885
#SPJ1
