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Salsk061 [2.6K]
3 years ago
11

You throw a bowling ball down an alley with speed v_0v ​0 ​​ , and initially it slides without rolling. After a little while fri

ction kicks in and the ball begins to roll. What is the speed of the bowling ball when it starts rolling without slipping?
Physics
1 answer:
telo118 [61]3 years ago
4 0

Answer:

final speed after pure rolling is given as R\omega = \frac{5}{7}v_0

Explanation:

As we know that point of contact at ground is taken as reference then there is no external torque about this point on the ball

So we can use angular momentum conservation about this point

L_i = L_f

L_i = mv_0R

L_f = \frac{7}{5}mR^2\omega

so we have

mv_0R = \frac{7}{5}mR^2\omega

R\omega = \frac{5}{7}v_0

So final speed after pure rolling is given as R\omega = \frac{5}{7}v_0

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sara and tory are out fishing on the lake on a hot summer day when they both decide to go for a swim. sara dives off the front o
crimeas [40]

Here it is an application of Newton's III law

as we know by Newton's III law that every action has equal and opposite reaction

So here as we know that two boys jumps off the boat with different forces

from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 45 N

Now similar way we can say

from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 60 N

So here net force due to both jump on the boat is given by

F_{net} = F_1 - F_2

F_{net} = 60 - 45

F_{net} = 15 N

so boat will have net force F = 15 N in forward direction due to both jumps

3 0
3 years ago
A particle executes simple harmonic motion with an amplitude of 2.18 cm.
Bogdan [553]

Answer:

The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.

Explanation:

Maximum speed is  :

                          v (max) = Aω

Speed v at any displacement y is given by  

v^{2} = w^{2} (A^{2} - y^{2})   ........................................................  i

And,

               v = \frac{1}{2} v (max)  

          or,  2 × v = Aω     ....................................................   ii

Eliminating  ω from equations i and ii,

                       \frac{1}{4} A^{2}  w^{2}  =  w^{2}  ( A^{2}  - y^{2})

                     or, y^{2} =  (\frac{3}{4}) A^{2}  =(\frac{3}{4}) 2.18^{2}

                    or,  y =  3.56 cm.

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3 years ago
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Nat2105 [25]
The answer they are looking for is the last one. However the last two are technically correct but the third one would result in negative work.
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Answer:

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