The magnitude of the magnetic dipole moment of the bar magnet is 1.2 Am²
<h3>
Magnetic dipole moment of the bar magnet</h3>
The magnitude of the magnetic dipole moment of the bar magnet at distance from its axis is calculated as follows;
where;
- B is magnetic field
- m is dipole moment
- μ is permeability of free space
m = (4π x 0.1³ x 2.4 x 10⁻⁴)/(2 x 4π x 10⁻⁷)
m = 1.2 Am²
The complete question is below:
What is the magnitude of the magnetic dipole moment of the bar magnet from 0.1 m of its axis and magnetic field strength of 2.4 x 10⁻⁴ T.
Learn more about dipole moment here: brainly.com/question/27590192
#SPJ11
Answer:
elastic potential energy
You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.
Explanation:
Answer:
The correct option is;
C. 1,715 m
Explanation:
We are given the information from the group of teen at the City edge
Time of arrival of explosion sound = 5 s after sighting
Time of sighting explosion = 5 s before hearing the boom
Speed of sound in air ≈ 343 m/s
Speed of light = 299,792 km/s
Therefore, distance covered by sound in 5 seconds is given by the following equation;
Hence Distance = 343 m/s × 5 s = 1715 m
To check, we compare the time it would take for the light to cover 1715 m
That is 
which is instantaneous hence the distance can be approximated by the time duration for the speed of sound.
Therefore, the distance of the students from the factory is approximately 1,715 m
Answer:
A) a = 73.304 rad/s²
B) Δθ = 3665.2 rad
Explanation:
A) From Newton's first equation of motion, we can say that;
a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.
Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s
Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s
We are given; t = 10 s
Thus;
a = 733.04/10
a = 73.304 rad/s²
B) From Newton's third equation of motion, we can say that;
ω² = ω_o² + 2aΔθ
Where Δθ is angular displacement
Making Δθ the subject;
Δθ = (ω² - ω_o²)/2a
At this point, ω = 0 rad/s while ω_o = 733.04 rad/s
Thus;
Δθ = (0² - 733.04²)/(2 × 73.304)
Δθ = -537347.6416/146.608
Δθ = - 3665.2 rad
We will take the absolute value.
Thus, Δθ = 3665.2 rad
Answer: 
Explanation:
The acceleration of an object can be calculated by using Newton's second law:
where
F is the net force applied on the object
m is the mass of the object
a is its acceleration
In this problem, we have F=125 N and m=25.0 kg, so we can rearrange the equation to calculate the acceleration:
