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otez555 [7]
2 years ago
9

Can anyone help me on this thnxs ​

Physics
1 answer:
DIA [1.3K]2 years ago
8 0

Answer:

a sin Ø - b cos Ø

_______________. =

a sin Ø+ b cos Ø

=> a - b cot ∅

_____________

a + b cot ∅

=> a - b x b

__

a

______________

a + b x b

___

a

=> a² - b ²

__________

a ² + b²

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In a 30.0-s interval, 500 hailstones strike a glass window with an area of 0.600m2 at an angle of 45.0°to the window surface. Ea
aliina [53]

or one hailstone we have;
Force = Mass X acceleration = 0.005kg x 9.8.} This is when the hailstone is not inclined at an angle.
When the hailstone is inclined at an angle of 45, then the component of force along the glass window will be F =0.005kg x 9.8 x sin45= 0.005kg x 9.8 x 0.707= 0.0346N.
Therefore, total force for the 500 hailstones would be 500x0.0346N=17.32N
This force is acting on an area equal to 0.600m2
Pressure = Force per unit area = 17.32N/0.600m2 = 28.9Pa

5 0
3 years ago
Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th
gregori [183]

Complete Question

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and thus 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 * 10^9years.

Answer:

The potassium-40 present in 80 kg is  Z = 0.0288 *10^{-3}\ kg

The effective dose absorbed per year is  x = 2.06 *10^{-24} per year

Explanation:

From the question we are told that

      The mass of potassium in 1 kg of human body is m =  3g= \frac{3}{1000} =  3*10^{-3} \ kg

      The mass of the person is M = 80 \ kg

       The abundance of Potassium-39 is   93.26%

        The abundance of Potassium-40 is   0.012%

         The abundance of Potassium-41 is   6.78 %

         The energy absorbed is  E =  1.10MeV = 1.10 *10^{6} * 1.602 *10^{-19} = 1.7622*10^{-13} J

Now  1 kg of human body contains       3.0*10^{-3}\ kg of  Potassium

So      80 kg of human body contains      k kg of  Potassium

=>   k = \frac{ 80 * 3*10^{-3}}{1}

     k = 0.240\  kg

Now from the question potassium-40 is  0.012% of the total  potassium so

     Amount of potassium-40  present is mathematically represented as

            Z = \frac{0.012}{100}  * 0.240

            Z = 0.0288 *10^{-3}\ kg

The effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is mathematically evaluated as

           D =  \frac{E}{M}

Substituting values

          D =  \frac{1.7622*10^{-13}}{80}

            D =  2.2*10^{-15} J/kg

Converting to Sieverts

We have

           D_s = REB * D

           D_s = 1.2 * 2.2 *10^{-15}

           D_s =  2.64 *10^{-15}

So

     for half-life (1.28 *10^9 \ years)  the dose is  2.64 *10^{-15}

     Then for 1  year the dose would be  x

=>         x = \frac{2.64 *10^{-15}}{1.28 * 10^9}

             x = 2.06 *10^{-24} per year      

7 0
3 years ago
A car travels 554 miles south in 10.4 hours . what is the velocity of the car ? show all steps
kirill115 [55]

Answer:

<h3>The answer is 53.27 mi/hr</h3>

Explanation:

To find the velocity covered by the car we use the formula

v =  \frac{d}{t}  \\

where

d is the distance

t is the time

From the question

d = 554 miles

t = 10.4 hrs

We have

d =  \frac{554}{10.4}  \\  = 53.2692307...

We have the final answer as

<h3>53.27 mi/hr</h3>

Hope this helps you

4 0
3 years ago
Help me with the following problem
timofeeve [1]

The rope will remain taut until the particle makes 79⁰ angle.

<h3>Change in kinetic energy of the particle</h3>

The change in kinetic energy of the particle is calculated as follows;

ΔK.E = K.Ei - K.Ef

Before the particle will achieve the given angular displacement, it will touch two new corners. Total kinetic energy lost = 30%

ΔK.E = 100%K.E - 30%K.E = 70%K.E = 0.7K.E

  • let the vertical displacement of the particle = h
  • horizontal length = side of the prism = a
  • hypotenuse side  = length of the pendulum = L
<h3>Apply principle of conservation of energy</h3>

K.E = P.E

0.7K.E = mgh

0.7(¹/₂mv²) = mg(Lsinθ)

0.7(v²) = 2g(Lsinθ)

from third kinematic equation;

v² = u² + 2gh

v² = 0 + 2gh

v² = 2g(a tanθ)

0.7(2g(a tanθ)) = 2g(Lsinθ)

0.7(a tanθ) = Lsinθ

0.7a/L = sinθ/tanθ

0.7a/L = cosθ

(0.7 x 0.8)/(3) = cosθ

0.1867 = cosθ

θ = cos⁻¹(0.1867)

θ = 79⁰

Thus, the rope will remain taut until the particle makes 79⁰ angle.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

5 0
1 year ago
Quarks
Basile [38]
C. T,B,U,D,C, and S. D is a tempting answer but it's actually the gluons that transmit the strong force.
4 0
2 years ago
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