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2 years ago

Carlos gets tired of pushing and instead begins to pull with force Fpull at an angle to the horizontal.

Physics

1 answer:

Pepsi [2]2 years ago

8 0

Answer:

The wrong items are;

1) The normal for FN equals the weight Fmg

2) The force of friction, Ff, equals the applied force Fpull

The corrected statements are;

1) The normal force is weight less the vertical component of the applied force Fpull

FN = Fmg - Fpull × sin(θ)

2) The force of friction equals the horizontal component of the applied force Fpull

Ff = Fpull × cos(θ)

Explanation:

The given statement was;

The velocity of the block is constant, so the net force exerted on the block must be zero. Thus, the normal force FN equals the weight Fmg, and the force of friction Ff equals the applied force Fpull

By the equilibrium of forces actin on the system, given that the applied force acts at an angle, θ, with the horizontal, we have;

The normal force is equal to the weight less the vertical component of the applied force;

That is we have, FN = Fmg - Fpull × sin(θ)

The friction force similarly, is equal to the horizontal component of the applied force;

Ff = Fpull × cos(θ)

The wrong items are therefore as follows;

1) The normal for FN equals the weight Fmg

1 i) The normal force is weight less the vertical component of the applied force Fpull

FN = Fmg - Fpull × sin(θ)

2) The force of friction, Ff, equals the applied force Fpull

2 i) The force of friction equals the horizontal component of the applied force Fpull

Ff = Fpull × cos(θ).

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A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu

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Answer:

please read the answer below

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The angular momentum is given by

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By taking into account the angles between the vectors r and v in each case we obtain:

v=(2,0)

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r=(1,0)

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angle = 180°

r and v are parallel

L = 0kgm/s

r=(1,1)

angle = 45°

$L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}$

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L = 5kgm/s

r=(-1,-1)

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$\Delta t = 2.0 ms = 0.002 s$ is the time interval

We need to find the magnetic flux before and after. The magnetic flux is given by:

$\Phi_B = BA$

where

B is the magnetic field intensity

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The radius of the coil is r = 12.0 cm = 0.12 m, so its area is

$A=\pi r^2 = \pi (0.12 m)^2 = 0.045 m^2$

At the beginning, the magnetic field is

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so the flux is

$\Phi_i = B_i A = (1.5 T)(0.045 m^2)=0.068 Wb$

while after the removal of the coil, the magnetic field is zero, so the flux is also zero:

$\Phi_f = 0$

so the variation of magnetic flux is

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And substituting into (1) we find the average emf in the coil

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