Describe rotational motion

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

disa [49]

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:

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7 0

1 year ago

Which symbol is used for an alpha particle? a. β c. γ b. Δ d. α

xenn [34]

It is the very last answer

6 0

2 years ago

A force of 6.00 N acts in the positive direction on a 3.00 kg object, originally traveling at +15.0 m/s, for 10.0 s. (a) What is

frozen [14]

Answer:

60 kg m/s

Explanation:

Let $a\;\; m/s^2$ be the acceleration of the object.

As the acceleration of the object is constant, so

$a=\frac {v-u}{t}\cdots(i)$

Given that applied force, F=6.00 N,

From Newton's second law, we have

$F= m\times a$ ,

$\Rightarrow F=\frac {m(v-u)}{t}$ [from equation (i)]

$\Rightarrow Ft=m(v-u)$

$\Rightarrow Ft=mv-mu$

$\Rightarrow mv-mu=6\times 10$ [given that time, t=10 s and F=6 N]

$\Rightarrow mv-mu=60 kg \;m/s$

Here mv is the final momentum of the object and mu is the initial momentum of the object.

So, the change in the momentum of the object is mv-mu.

Hence, the change in the momentum of the object is 60 kg m/s.

6 0

2 years ago

A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2. Compu

boyakko [2]

Answer:

0.42 m/s²

Explanation:

r = radius of the flywheel = 0.300 m

w₀ = initial angular speed = 0 rad/s

w = final angular speed = ?

θ = angular displacement = 60 deg = 1.05 rad

α = angular acceleration = 0.6 rad/s²

Using the equation

w² = w₀² + 2 α θ

w² = 0² + 2 (0.6) (1.05)

w = 1.12 rad/s

Tangential acceleration is given as

$a_{t}$ = r α = (0.300) (0.6) = 0.18 m/s²

Radial acceleration is given as

$a_{r}$ = r w² = (0.300) (1.12)² = 0.38 m/s²

Magnitude of resultant acceleration is given as

$a = \sqrt{a_{t}^{2} + a_{r}^{2}}$

$a = \sqrt{0.18^{2} + 0.38^{2}}$

$a$ = 0.42 m/s²

8 0

2 years ago

Unpolarizedlight of intensity I_0 is incident on three polarizingfilters. The axis of the first is vertical, that of the secondi

Marina86 [1]

To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:

$I = I_0 cos^2\theta$