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natali 33 [55]
3 years ago
11

A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho

rizontal floor. Neglect any friction acting on the piano.Calculate the magnitude of the force applied by the man if he pushes
(a) parallel to the incline and
(b) parallel to the floor.

Physics
1 answer:
creativ13 [48]3 years ago
6 0

Answer:

The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.

Explanation:

Given:

Mass of the piano (m) = 190 kg

Inclined angle (\theta) = 18 degree

Considering gravity, g = 9.8 ms^-^2

And

Using, sin(18) =0.30 and cos(18)=0.95

<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

a.

When the man pushes it parallel to the incline.

Balancing the forces as  \sum F=0 .

⇒ F+mgsin(\theta) =0

⇒ F=-mgsin(\theta)

⇒ Here it is negative as the force is acting downward.

⇒ Plugging the values of mass (m) and angle (\theta) .

⇒ F=190\times 9.8\times sin(18)

⇒ F=575.38 N

b.

When the force is parallel to the floor.

⇒ Fcos(\theta)=mgsin(\theta)

⇒ F=\frac{mgsin(\theta)}{cos(\theta)}

⇒ Plugging the values.

⇒ F=\frac{190\times 9.8\times sin(18)}{cos(18)}

⇒ F=605 N

So,

The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

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

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      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}

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     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      

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Learn more about forces and weight:

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