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zysi [14]
3 years ago
5

If the caffeine concentration in a particular brand of soda is 2.97 mg/oz, 2.97 mg/oz, drinking how many cans of soda would be l

ethal? Assume that 10.0 g of caffeine is a lethal dose, and there are 12 oz in a can.
Physics
1 answer:
Ray Of Light [21]3 years ago
5 0

Explanation:

The given data is as follows.

     Concentration of caffeine = 2.97 mg/oz

     Number of oz in a can = 12 oz

Therefore, the concentration of caffeine in one can is calculated as follows.

                 = (12 \times 2.97) mg

                 = 35.64 mg

                 = 35.64 \times 10^{-3} g

Since, it is given that lethal dose is 10.0 g. Hence, number of cans are calculated as follows.

     No. of cans = \frac{\text{Lethal dose}}{\text{concentration in one can}}

                         = \frac{10 g}{35.64 \times 10^{-3} g}

                         = 280.58

                         = 281 (approx)

Thus, we can conclude that 281 cans of soda would be lethal.

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A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes
OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

y(t) = h +u_y t - \frac{1}{2}gt^2

where

h = 2.5 km = 2500 m is the initial height

u_y = 0 is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

y(t) = 0

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s

So the horizontal distance travelled is

d=v_x t

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

d=(333.3)(22.6)=7533 m = 7.5 km

7 0
3 years ago
Bill and Nageen have each built an electric motor and they are testing them by having them lift boxes. Bill's motor lifts a box
photoshop1234 [79]

Answer:

Bill's motor power: W_B = F x S / T = F x 0.35 / 2= 0.175F

Nageen's motor power: W_N = F x S / T = F x 0.35 / 1.8 = 0.194F

=> 0.194F > 0.175F => Nageen's motor applied more power to the box than Bill's motor.

5 0
3 years ago
An automobile rounds a curve of radius 50.0 m on a flat road.
bixtya [17]

Answer:

14m/s

Explanation:

Given parameters:

Radius of the curve  = 50m

Centripetal acceleration  = 3.92m/s²

Unknown:

Speed needed to keep the car on the curve = ?

Solution:

The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.

 It is given as;

      a = \frac{v^{2} }{r}  

a is the centripetal acceleration

v is the speed

r is the radius

  Now insert the parameters and find v;

         v²   = ar

        v² = 3.92 x 50  = 196

         v  = √196 = 14m/s

6 0
3 years ago
What is the first semiconductor
madam [21]
<span>
English "natural philosopher" (the contemporary term for physicist) Michael Faraday is renowned for his discovery of the principles of electro-magnetic induction and electro-magnetic rotation, the interaction between electricity and magnetism that led to the development of the electric motor and generator. The unit of measurement of electrical capacitance - the farad (F) - is named in his honor. Faraday's experimental work in chemistry, which included the discovery of benzene, also led him to the first documented observation of a material that we now call a semiconductor. While investigating the effect of temperature on "sulphurette of silver" (silver sulfide) in 1833 he found that electrical conductivity increased with increasing temperature. This effect, typical of semiconductors, is the opposite of that measured in metals such as copper, where conductivity decreases as temperature is increased. In a chapter entitled "On Conducting Power Generally" in his book Experimental Researches in Electricity Faraday writes "I have lately met with an extraordinary case ... which is in direct contrast with the influence of heat upon metallic bodies ... On applying a lamp ... the conducting power rose rapidly with the heat ... On removing the lamp and allowing the heat to fall, the effects were reversed." We now understand that raising the temperature of most semiconductors increases the density of charge carriers inside them and hence their conductivity. This effect is used to make thermistors - special resistors that exhibit a decrease in electrical resistance (or an increase in conductivity) with an increase in temperature. <span> Next Milestone </span> Contemporary Documents <span> <span>Faraday, M. Experimental Researches in Electricity, Volume 1. (London: Richard and John Edward Taylor, 1839) pp.122-124 (para. 432). Note: This section appears on different pages in later editions of the book. The material in the book is reprinted from articles by Faraday published in the Philosophical Transactions of the Royal Society of 1831-1838. </span> </span> More Information <span> <span>Hirshfeld, Alan W. The Electric Life of Michael Faraday. Walker & Company (March 7, 2006).</span> <span>Friedel, Robert D. Lines and Waves: Faraday, Maxwell and 150 Years of Electromagnetism. Center for the History of Electrical Engineering, Institute of Electrical and Electronics Engineers (1981).</span> </span> </span>
3 0
4 years ago
Three point charges are placed on the y-axis: a charge q at y=a, a charge –2q at the origin, and a charge q at y= –a. Such an ar
den301095 [7]

Answer:

electric field   Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

Explanation:

The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.

 a) Let's write the electric field for each charge and the total field

       E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

       Et = E1 + E2 + E3

       E1 = k q / (x-a)²

       E2 = k (-2q) / x²  

       E3 = k q / (x + a)²

       Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ  where x << 1, for this we take factor like x from all the equations

       Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

     (1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1

     (1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

       

   Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

   Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]

Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

3 0
3 years ago
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