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Svetllana [295]
3 years ago
8

A particle of mass m=375 g is launched with velocity of A =4 moves with a length AB=2.5m under the action of tractive force F=3.

46 N making an angle 30. during its motion between A and B which is subjected to a frictional force f=1.5 N
calculate V of B by applying the kinetic energy theorem​
Physics
1 answer:
DiKsa [7]3 years ago
3 0

Answer:

The final speed is 5.78 m/s.

Explanation:

mass, m = 375 g = 0.375 kg

initial velocity, u = 4 m/s

Distance, s = 2.5 m

Angle, A = 30 degree

Force, F = 1.5 N

let the final velocity is v.

Use the work energy theorem

Work done = change in kinetic energy

W= 0.5 m(v^2 - u^2)\\\\F s cos A= 0.5 m (v^2 - u^2)\\\\1.5\times 2.5\times cos30= 0.5\times 0.375\times (v^2 - 16)\\\\v = 5.78 m/s

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Answer:

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d\vec{A} is parallel to \vec{B} and \vec{B} is constant in magnitude and direction therefore:

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Part A)

initially the flux is \phi =\pi B r^2

after the interval \Delta t= 2.4 [m/s]

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now, the EMF is defined as:

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if we consider \Delta t= 2.4 [m/s] very small then we can re-write it as:

\epsilon =- \frac{\Delta \phi}{\Delta t}

\Delta \phi = 0 - \pi B r^2=-\pi (1.7) (0.15)^2=-0.12

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Part B)

When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.

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Serggg [28]

Answer:

<em>n =1.33 revolutions</em>

Explanation:

<u>Uniform Circular Motion</u>

The angular speed can be calculated in two different ways:

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