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Colt1911 [192]
3 years ago
13

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th

us our bodies) as three isotopes: Potassium-39, Potassium-40, and Potassium-41. Their current abundances are 93.26%, 0.012% and 6.728%. A typical human body contains about 3.0 grams of Potassium per kilogram of body mass. 1. How much Potassium-40 is present in a person with a mass of 80 kg? 2. If, on average, the decay of Potassium-40 results in 1.10 MeV of energy absorbed, determine the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body. Assume an RBE of 1.2. The half-life of Potassium-40 is 1.28 x 10°years.
Physics
1 answer:
andriy [413]3 years ago
3 0
Just grab a calculator
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What is the pressure drop due to the bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire
Paul [167]

Answer:

\Delta P=1581357.92\ Pa

Explanation:

Given:

  • diameter of hose pipe, D=0.09\ m
  • diameter of nozzle, d=0.03\ m
  • volume flow rate, \dot{V}=40\ L.s^{-1}=0.04\ m^3.s^{-1}

<u>Now, flow velocity in hose:</u>

v_h=\frac{\dot V}{\pi.D^2\div 4}

v_h=\frac{0.04\times 4}{\pi\times 0.09^2}

v_h=6.2876\ m.s^{-1}

<u>Now, flow velocity in nozzle:</u>

v_n=\frac{\dot V}{\pi.d^2\div 4}

v_n=\frac{0.04\times 4}{\pi\times 0.03^2}

v_n=56.5884\ m.s^{-1}

We know the Bernoulli's equation:

\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}+Z_2

when the two points are at same height then the eq. becomes

\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}

\Delta P=\frac{\rho(v_n^2-v_h^2)}{2}

\Delta P=\frac{1000(56.5884^2-6.2876^2)}{2}

\Delta P=1581357.92\ Pa

8 0
3 years ago
An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 49.8 N,
tatiyna

Answer:

W = 7.06 J

Explanation:

From the given information the spring constant 'k' can be calculated using the Hooke's Law.

F = kx\\49.8 = k(0.181)\\k = 275.13~N/m

Now, using this spring constant the additional work required by F to stretch the spring can be found.

The work energy theorem tells us that the work done on the spring is equal to the change in the energy. Therefore,

W = U_2 - U_1\\W = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 = \frac{1}{2}(275.13)[0.29^2 - 0.18^2] = 7.06~J

6 0
3 years ago
Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha
Veronika [31]

Answer:

The force of friction acting on block B is approximately 26.7N.  Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.  

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation  is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

a_B = a_A-0.5 \frac{m}{s^2}

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

6 0
3 years ago
Read 2 more answers
Need help 8th grade science test need help
jolli1 [7]
ANSWER: d, average speed
7 0
2 years ago
Read 2 more answers
An object with a charge of 0.9 x 10^-5 C is separated from a second object with a chare of 2.5 x 10^-4 C by a distance of 0.5 m.
meriva

Answer: Force = 81 N

Explanation:

from Columbs law,

F = k(q1*q2)/r²

k = 9 x 10^9 Nm²/C²

F = (9 x 10^9)x (0.9x10^-5 x 2.5x10^-4)/(0.5)²

F = 81 Newtons

4 0
3 years ago
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