<u>Question:</u>
You are working on an experiment involving a very strong permanent magnet, and your data suggests that your magnet's field suddenly decreased during some interval in time. Such a decrease could have been caused by the magnet
A. Having overheated substantially
B. Being hit hard
C. Both A and B
D. Being grounded out
<h3><u>Answer:</u></h3>
A decrease in magnetic field of the permanent magnet have been caused by the magnet having overheated substantially or sharp impacts by being hit hard.
Option c
<h3><u>Explanation: </u></h3>
Permanent magnets are ferromagnetic materials with its magnetic domains aligned and grouped together in the same direction. These atomic domains maintain their directionality and hence a permanent magnet provides persistently strong magnetic fields without quick weakening. Some factors may lead to demagnetization or else a consistent reduction in magnetic strength.
Overheating a magnetic material realigns the magnetic domain regions and affects its directionality. When it reaches to a temperature defined as Curie temperature, varying with each material; the substance is no more a magnet due to complete randomness in the domain structure. As the temperature decreases and approaches the room temperature, magnetic field appears but is less in strength. Sudden impacts due to hitting may lead to random realignment of magnetic domains and thus decrease its magnetic strength.
<span>Mass of the block m = 3.3kg
Angle of the slide = 30 degrees
Distance the block slides s = 2.10 m
Time taken to slide t = 1.6 s
Initially in rest condition so initial velocity u = 0.
We have an equation for distance s = (u x t) + (1/2) x (a t^2)
s = (0 x t) + (1/2) x (a x (1.6) ^2) => 2.10 = (1/2) x (a x2.56)
2.56a = 4.20 => a = 1.64
So the magnitude of the Acceleration a = 1.64 m/s^2</span>
The magnitude of the force is F=1.68×10 ^20 N
<u>Explanation:</u>
<u>Given data</u>
q1 =1.5 10 ^6 q2=3.2 10^6 r=1.5
<u>We have the formula</u>
By the coulomb's law
F= K. q1 ×q2 / r²
The K value is given by
8.99 10^9 Nm²/ c²
substitute the values we get,
F= ( 8.99× 10^9 Nm²/ c²) ×(<u>1.5 ×10 ^6</u>)×(<u>3.2 ×10 ^6</u>)/ (1.6 m² )
F=1.68×10 ^20 N
The magnitude of the force is F=1.68×10 ^20 N
<span>8,480 Joules. You must multiply by the angle in radians through which the target object rotates. 4.5 x 2 x 3.14 = 28.26666666 J for each turn
28.26666666 x 300 = 8480</span>
Answer:
Explanation:
F=ma
F in this case is the gravity acting on the 2kg object. Acceleration of gravity is 9.8 m/s^2. SF=ma, so
F = 2kg*(9.8 m/s^2) = 19.6 N
Now use this force to determine the mass of the object on the table:
F=ma
19.6 N (1N=kg*m/s^2) = m*(1.8 m/s^2)
m = 10.89 kg
