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Harrizon [31]
3 years ago
15

The half-life of caffeine is 5 hours. If you ingested a 30 oz Big Gulp, after how much time will have to pass before you have un

der 1 oz of caffeine remaining?
Physics
1 answer:
VladimirAG [237]3 years ago
6 0

Answer:

The time that will have to pass before one has under 1 oz of caffeine remaining is 24.53 hours

Explanation:

Here, we have the formula for half life as follows;

N(t) = N_0(\frac{1}{2})^{\frac{t}{t_{1/2}}

Where:

N(t) = Remaining quantity of the substance = 1 oz

N₀ = Initial quantity of the substance = 30 oz

t = Time duration

t_{1/2} = Half life of the substance = 5 hours

Therefore, plugging in the values, we have

1= 30(\frac{1}{2})^{\frac{t}{5}}

\frac{1}{30} = (\frac{1}{2})^{\frac{t}{5}}\\ln(\frac{1}{30}) =\frac{t}{5} ln(\frac{1}{2})\\\frac{t}{5}  = \frac{ln(\frac{1}{30}) }{ ln(\frac{1}{2})} = 4.91\\ t = 4.91 \times 5 = 24.53 \ hours

The time that will have to pass before one has under 1 oz of caffeine remaining = 24.53 hours.

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A simple generator is used to generate a peak output voltage of 19.0 V . The square armature consists of windings that are 6.65
salantis [7]

One of the efficient concepts that can help us find the number of turns of the cable is through the concept of induced voltage or electromotive force given by Faraday's law. The electromotive force or emf can be described as,

\epsilon = NBA\omega

Where,

N = Number of loops

B = Magnetic Field

A = Cross-sectional Area

\omega = Angular velocity

Re-arrange to find N,

N = \frac{\epsilon}{BA\omega}

Our values are given as,

\epsilon = 19V

B = 0.434T

\omega = 49.8\frac{rev}{s} (\frac{2\pi rad}{1 rev}) = 99.6\pi rad/s

A = (6.65*10^{-2})^2 m^2

Replacing at our equation we have:

N = \frac{\epsilon}{( 0.434)A\omega}

N = \frac{19}{( 0.434)((6.65*10^{-2})^2)(99.6\pi)}

N = 31.63 \approx 32

Therefore the number of loops of wire should be wound on the square armature is 32 loops

6 0
4 years ago
A 65kg person throw a 0.045kg snowball forward with a ground speed of 30m/s. A second person, with a mass of 60kg, catches the s
Kobotan [32]
Well, st first we should find <span>initial momentum for the first person represented in the task which definitely must be :
</span>(65+0.045)*2.5
And then we find the final one :  65*x + 0.045*30
Then equate them together : x=2.48 m/s 
So we can get the velocity, which is is 2.48 m/s
In that way, according to the main rules of <span>conservation of momentum you can easily find the solution for the second person.
Regards!</span>
6 0
4 years ago
I need help please, thank you!!!
stiks02 [169]

Answer:

135 mph

because 7+9=16

8 is left

if 7=40

9=50

then 8=45

add 40+50+45mph

=135

4 0
3 years ago
In a college homecoming competition, eighteen students lift a sports car. While holding the car off the ground, each student exe
Nata [24]

Answer:

Explanation:

Given

Each student exert a force of F=400 N

Let mass of car be m

there are 18 students who lifts the car

Total force by 18 students F=18\times 400=7200 N

therefore weight of car W=7200

mass of car m=\frac{W}{g}

m=\frac{7200}{9.8}=734.69 kg

(b)7200 N \approx 1618.624\ Pound-force

734.69 kg\approx 1619.71 Pounds                  

6 0
3 years ago
A bicyclist is travelling at 25 m/s when he begins to decelerate at -4m/s2 . How fast is travelling after 5 seconds
Umnica [9.8K]

Initial velocity (Vi) = 25 m/s

acceleration (a) = -4 m/s^{2}

time interval (t) = 5 sec

let us assume that final velocity after 5 sec be Vf

As acceleration is constant, we can apply the the equation of motion with constant acceleration i.e. V_{f} = V_{i} + at

Hence, V_{f} = 25 +(-4)(5) = 25 -20 = 5 m/s

so, the velocity of bicyclist will be 5 m/s after 5 sec

7 0
3 years ago
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