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Andrew [12]
3 years ago
15

The half-life of caffeine is 5 hours. If you ingested a 30 oz Big Gulp, how many oz of caffeine is left after one half life? * Y

our answer
Physics
1 answer:
xeze [42]3 years ago
8 0

Answer:

The amount of caffeine left after one half life of 5 hours is 15 oz.

Explanation:

Half life is the time taken for a radioactive substance to degenerate or decay to half of its original size.

The half life of caffeine is 5 hours. So ingesting a 30 oz, this would be reduced to half of its size after the first 5 hours.

So that:

After one half life of 5 hours, the value of caffeine that would be left is;

                                    \frac{30}{2} = 15 oz

The amount of caffeine left after one half life of 5 hours is 15 oz.

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An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the
OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2

Through the aforementioned formula we will have to

v_f^2-v_i^2 = 2ax

The particulate part of the rest, so the final speed would be

v_f^2 = 2gx

v_f=\sqrt{2(9.8)(3.1)}

v_f = 7.79m/s

Now from Newton's second law we know that

F = ma

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

F = m\frac{dv}{dt}

Replacing we have that,

F = (89)\frac{7.79}{0.5}

F = 1386.62N

Therefore the force that the water exert on the man is 1386.62

3 0
3 years ago
Electrical energy can be transformed into other types of energy. We often experience this transformation of energy in our everyd
vitfil [10]
Well I’m not sure because you don’t have anything listed
3 0
3 years ago
What is the frequency of an electromagnetic wave that has a wavelength of 300,000 km? (the speed of light is 300,000 km/s.)?
expeople1 [14]
Frequency represents the number of complete oscillations in one second. it is measured in Hertz (Hz). Electromagnetic waves are waves which do not require a material media for transmission. They travel with a speed of light.
The speed (m/s) of a wave is given by  frequency (Hz) × Wavelength (m)
Speed is 300,000 km/sec or 300,000,000 m/s and the wavelength is 300,000  km or 300,000,000 m.
Frequency = speed÷ wavelength
                 = 300000000 ÷ 300000000 = 1
Therefore, the frequency of the wave is 1Hz

6 0
3 years ago
Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat
Alja [10]

Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

Note: the mass doesn't change, it is only the velocity that change.

Also,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Let assume that if is constant acceleration

Then, m and a is constant,

Then,

K.E is directly proportional to d

So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

K.E_1 = E_k

d_1 = d

d_2 = d/3

K.E_2 = K.E_1 / d_1 × d_2

K.E_2 = E_k × ⅓d / d

Then,

K.E_2 = ⅓E_k

So, the new kinetic energy is one third of the E_k

7 0
3 years ago
The atmosphere of Mars is almost all carbon dioxide and the average surface pressure is 610 Pa (as compared with 101,000 Pa on E
Karolina [17]

Answer:

   z = 3,737 10⁵ m

Explanation:

a) As they indicate that the atmosphere behaves like an ideal gas, we can use the equation

          P V = n R T

          P = (n r / V) T

We replace

         P = (n R / V) T₀ e^{- C z}

b) Let's apply this equation in the points

Lower

        .z = 0

         P₀ = 610 Pa

         P₀ = (nR / V) T₀

Higher.

         P = 10 Pa

          P = (n R / V) T₀ e^{- C z}

We replace

        P = P₀ e^{- C z}

        e^{- C z} = P / P₀

        C z = ln P₀ / P

        z = 1 / C ln P₀ / P

Let's calculate

        z = 1 / 1.1 10⁻⁵ ln (610/10)

        z = 3,737 10⁵ m

4 0
3 years ago
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