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polet [3.4K]
2 years ago
11

A ball of mass m kg is rolled up a slope at an angle a to the horizontal, where sin a = 2/5. The ball passes a point A with spee

d 7ms.-1
A point B is 5 m further up the slope than point A. Find the time between passing
B on the way up and returning to B on the way down.
Physics
1 answer:
myrzilka [38]2 years ago
5 0

16.42 seconds will elapse between passing B on the way up and coming back to B on the way down.

<h3>What is the definition of Angular Speed? </h3>

The rate of change of angular displacement is defined as angular speed, and it is stated as follows:

ω = θ t

Where,

θ is the angle of rotation,

t is the time

ω is the angular velocity

Given data;

sin a = 2/5

a=sin(2/5)

a=23°

The linear velocity is the product of the angular velocity and the length.

\rm v= \omega r \\\\ \omega= 7/5 \\\\ \omega= 1.4 \ rad/sec

The times between the two points are found as;

\omega = \frac{\triangle \theta }{\triangle t} \\\\ t= \frac{23}{1.4} \\\\ t = 16.42 \ sec

Hence, the time between passing B on the way up and returning to B on the way down will be 16.42 sec.

To learn more about the angular speed refer to the link;

brainly.com/question/9684874

#SPJ1

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An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl
svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 0 m/s

         Displacement, s = 192 m

         Acceleration, a = 9.81 m/s²

         Substituting

                      s = ut + 0.5 at²

                      192 = 0 x t + 0.5 x 9.81 x t²

                         t = 6.26 seconds

         Now we need to find horizontal distance traveled in 6.26 seconds by the package.

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 41 m/s

        Time, t = 6.26 s

         Acceleration, a = 0 m/s²

         Substituting

                      s = ut + 0.5 at²

                      s = 41 x 6.26 + 0.5 x 0 x 6.26²

                         s = 256.52 m

     The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

          Initial velocity, u = 0 m/s

         Time, t = 6.26 s

         Acceleration, a = 9.81 m/s²  

         Substituting

                      v = u+ at

                       v = 0 + 9.81 x 6.26 = 61.41 m/s

   Vertical component of velocity = 61.41 m/s      

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3 years ago
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