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Serggg [28]
3 years ago
9

A flea jumps by exerting a force of 1.07 10-5 N straight down on the ground. A breeze blowing on the flea parallel to the ground

exerts a force of 1.14 10-6 N on the flea. Find the direction and magnitude (in m/s2) of the acceleration of the flea if its mass is 6.0 10-7 kg. (Let us assume that).
Physics
1 answer:
Aleonysh [2.5K]3 years ago
8 0

Answer:

a) 15.77 m/sec2

b) 13.3 deg

Explanation:

we are given;

Flea force = F1=1.07×10⁻5 N j

Breeze force = F2 = 1.14× 10⁻6 N (-j

mass of flea =6.0 ×10⁻7 kg

So net force on the flea=F1+F2+weight of flea=1.07×10⁻5 j +1.14× 10⁻6 i + 6.0 ×10⁻7 (-j) ×9.8= ma

==> ma = 1.07×10⁻5 j - 0.588×10⁻5 j + 0.114×10⁻5 i

==> ma= 0.482 ×10⁻5 j +0.114×10⁻5 i

==> ma = 0.114×10⁻5 i +0.482 ×10⁻5 j

== a = (0.114×10⁻5 i +0.482 ×10⁻5 j) / 6.0 ×10⁻7

==> a =

==>a= (1.9 j+8.03 i ) m/sec2

mag of a \sqrt{1.9^{2}  +8.03^{2} = 15.77 m/sec2

direction angle = tan⁻1(1.9/8.03)=13.3°

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A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th
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Answer:

The height is  h_c = 42.857

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

   From the question we are told that

           The height is h_s = 30 \ cm

            The angle of the slope is \theta = 15^o

According to the law of conservation of energy

     The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

                          mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2

Where I is the moment of inertia which is mathematically represented as this for  a sphere

                    I = \frac{2}{5} mr^2

  The angular velocity w is mathematically represented as

                         w = \frac{v}{r}

So the equation for conservation of energy becomes

               mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2

              mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]

             mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]

            gh_s =[\frac{7}{10} ] v^2

              v^2 = \frac{10gh_s}{7}

Considering a circular hoop

   The moment of inertial is different for circle and it is mathematically represented as

             I = mr^2

Substituting this into the conservation equation above

              mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2

Where h_c is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

                 mgh_c  = mv^2

                     gh_c = v^2

                     h_c = \frac{v^2}{g}

Recall that   v^2 = \frac{10gh_s}{7}

                    h_c= \frac{\frac{10gh_s}{7} }{g}

                      = \frac{10h_s}{7}

            Substituting values

                   h_c = \frac{10(30)}{7}

                       h_c = 42.86 \ cm    

       

     

                         

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