D-It will become a temporary magnet because the domains will easily realign.
Answer:
Pluto
Explanation:
In our solar system, we have several planet. Pluto is one of the. Pluto is a planet that is highly oval shaped orbit and eccentric that brings it inside the another orbit. It get inside the orbit of Neptune. Sometimes even Neptune get far away from sun in comparison to the dwarf planet Pluto.
It is very strange happening in the world of planet. it occurs in the year of 1979 and 1999. But Pluto never ever crashed into Neptune. It happen because Neptune takes every three lapse that takes around the sun but Pluto makes only two lapse. This happening prevents two bodies from clashes.
Answer:
Explanation:
*Assume the parallel disks have equal diameters.
Given the electric strength as 
transferring 
electrons, the disk's Area can be calculated using the formula:
#We now calculate the disks diameter:
Hence, the diameter of the disks is
Mass = 1kg
Distance = 1m
Time = 1s
Force= Mass x Acceleration due to graviy
= 1 x 9.8 = 9.8
Velocity = Distance / time
= 1 / 1 =1m/s
Power = Force x velocity
= 9.8 x 1 = 9.8 W
<span>1.57 seconds.
The rod hanging from the nail constructs a physical pendulum. The period of such a pendulum follows the formula
T = 2*pi*sqrt(L/g)
where
T = time
L = length of pendulum
g = local gravitational acceleration
So the problem becomes one of determining L. It's tempting to consider L to be the distance between the center of mass and the pivot, but that isn't the right value. The correct value is the distance between the pivot and the center of percussion. So let's determine what that is. We can treat the uniform thin rod as an uniform beam and for an uniform beam the distance between the center of mass and the center of percussion is expressed as
b = L^2/(12A)
where
b = distance between center of mass and center of percussion
L = length of beam
A = distance between pivot and center of mass
Since the rod is uniform, the CoM will be midway from either end, or 0.962 m / 2 = 0.481 m from the end. The pivot will therefore be 0.481 m - 0.048 m = 0.433 m from the CoM
Now let's calculate the distance the CoP will be from the CoM:
b = L^2/(12A)
b = (0.962 m)^2/(12 * 0.433 m)
b = (0.925444 m^2)/(5.196 m)
b = 0.178107005 m
With the distance between the CoM and CoP known, we can now calculate the effective length of the pendulum. So:
0.433 m + 0.178107005 m = 0.611107005 m
And finally, with the effective length known, let's calculate the period.
T = 2*pi*sqrt(L/g)
T = 2*pi*sqrt((0.611107005 m)/(9.8 m/s^2))
T = 2*pi*sqrt(0.062357858 s^2)
T = 2*pi*0.249715554 s
T = 1.569009097 s
Rounding to 3 significant figures gives 1.57 seconds.
Let's check if this result is sane. Looking up "Seconds Pendulum", I get a length of 0.994 meters which is longer than the length of 0.611 meters calculated. But upon looking closer at the "Seconds Pendulum", you'll realize that it's period is actually 2 seconds, or 1 second per swing. So the length of the calculated pendulum is sane.</span>
