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WITCHER [35]
2 years ago
14

A student walks 50 meters east, 40 meters north, 35 meters east, and then 20 m south. What is the magnitude and direction of the

student's total displacement
Physics
1 answer:
Nuetrik [128]2 years ago
3 0

Answer: A student walks 50 meters east, 40 meters north, 35 meters east, and then 20 m south. Then the magnitude and direction of the student's total displacement will be 87.32 m along the direction of AD or in east-south direction.

Explanation: To find the correct answer, we need to know about the Displacement of a body in motion.

<h3>What is displacement of a body in motion?</h3>
  • The displacement is the shortest distance between initial and final positions of a body.
  • It's a vector quantity, and can positive, negative, or zero.
  • The magnitude of displacement is less than or equal to the distance travelled.
<h3>How to solve the problem?</h3>
  • At first, we can draw a diagram showing the motion of the body.
  • From the diagram, the displacement of the body will be equal to the distance between point A and D.
  • To solve this, we can use Pythagoras theorem.

AD=AC+CD\\AC^{2} =50^{2} +11^2\\AC=51.19 m\\Similarly,\\CD^2=35^2+9^2\\CD=36.13 m\\thus, \\AD=51.19+36.13=87.32 m

Thus, from the above calculations, we can conclude that, the displacement of the body will be equal to 87.32 m along the direction of AD or in east-south direction.

Learn more about the Displacement here:

brainly.com/question/28020108

#SPJ4

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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
photoshop1234 [79]

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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So, Force * distance
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x² * x³ = x²⁺³ = x⁵

Therefore, It would be kgm²s⁻²
m¹ * m¹ = m¹⁺¹ = m²
s⁻² is also = s / 2





4 0
3 years ago
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