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1 year ago

If the bar magnet is flipped over and the south pole is brought near the hanging ball, the ball will be?

Physics

1 answer:

disa [49]1 year ago

7 0

The ball may attracted to the magnet.

<h3>How can we understand that the hanging ball will be attracted to the magnet or not?</h3>

From the question, we understand that the ball is attracted by the north pole of the bar magnet, then the bar magnet flipped over and the south pole is brought near the hanging ball.
As we know, in this type of experiments of bar magnet most of the times the ball is made out of steel.
Steel is a magnetic material.
Magnetic materials gets attracted to the magnet at both the North and South pole.
This can be compared to how neutral objects also gets attracted to the positively and negatively charged rods through the Polarization force.

So, If the bar magnet is flipped over and the south pole is brought near the hanging ball, The ball will be attracted to the magnet.

Learn more about the bar magnet:

brainly.com/question/27943723

#SPJ4

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3 years ago

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A historical society is testing an old cannon. They

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Answer:

26.2 m/s

Explanation:

We can find the speed of the cannonball just by analyzing its vertical motion. In fact, the initial vertical velocity is

$u_y = u sin \theta$ (1)

where u is the initial speed and $\theta = 45.0^{\circ}$ is the angle of projection.

We can therefore use the following suvat equation for the vertical motion of the ball:

$v_y = u_y + at$

where $v_y$ is the vertical velocity at time t, and $a=g=-9.8 m/s^2$ is the acceleration of gravity. The time of flight is 3.78 s, so we know that the ball reaches its maximum height at half this time:

$t=\frac{3.78}{2}=1.89 s$

And at the maximum height, the vertical velocity is zero:

$v_y=0$

Substituting these values, we find the initial vertical velocity:

$u_y = v_y - at = 0-(-9.8)(1.89)=18.5 m/s$

And using eq.(1) we now find the initial speed:

$u=\frac{u_y}{sin \theta}=\frac{18.5}{sin 45.0^{\circ}}=26.2 m/s$

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3 years ago

The primary cause of weather is due to ______

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Answer:

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3 years ago

A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed vcom,0 = 6.0 m/s

ankoles [38]

Answer:

Explanation:

Radius of the ball is $R=11cm=0.11m$

Initial speed of the ball is $v_{com0}=6.0m/s$

Initial angular speed of the ball is $\omega = 0$

Coefficient of kinetic friction between the ball and the lane

is $\mu =0.35$

Due to the presence of frictional force, ball moves with

decreasing velocity.

(a)

velocity $v_{com0}$ in terms of $\omega$ is

$V_{com0} = -R\omega\\\\=-(0.11m)\omega\\\\= (-0.11\omega)m/s$

(b)

Ball's linear acceleration is given by

$a=-\mu g\\\\=-(0.35) (9.8 m/s^2)\\\\= -3.43m/s^2$

(c)

During sliding, ball's angular acceleration is calculated as

$\alpha=-\frac{\tau}{I}\\\\-\frac{\mu mgR}{(\frac{2}{5}mR^2)}\\\\-\frac{2}{5}\frac{\mu g}{R}\\\\-\frac{2}{5}\frac{(0.35)(9.8)}{0.11}\\\\-77.95rad/s^2$

(d)

The time for which the ball slides is calculated from the

equation of motion is

$V_{cm}= V_{cm0} + at\\\\V_{cm} = V_{cm0} + (-\mu g )t\\\\-0.11\omega=6.0m/s -(3.43m/s^2 )t\\\\-(0.11) (\alpha t) =6.0 m/s - (3.43 m/s^2)\\\\- (0.11)(-77.95 rad/s^2)t = 6.0m/s - (3.43 m/s^2 )t\\\\8.5745t + 3.43t= 6.0\\\\12.0045t = 6.0\\\\t= 0.4998s$