Answer:
1)
Force on bar magnet = 0
Torque on bar magnet = 0
2)
Force on bar magnet = 0
Torque on bar magnet = 0.177 Nm
3)
Force on bar magnet = 0
Torque on bar magnet = 0.25 Nm
Explanation:
Part 1)
net force on bar magnet in uniform magnetic field is always zero
Torque on bar magnet is given as

when bar magnet is inclined along z axis along magnetic field
then we will have

Part 2)
net force on bar magnet in uniform magnetic field is always zero
Torque on bar magnet is given as

when bar magnet is pointing 45 degree with z axis then we will have



Part 3)
net force on bar magnet in uniform magnetic field is always zero
Torque on bar magnet is given as

when bar magnet is pointing 90 degree with z axis then we will have



Explanation:
Given that, the height of the tide measured at a seaside community varies according to the number of hours t after midnight. The height is given by the equation as :

When the tide first be at 6 ft, put h = 6 ft in above equation as :


On solving the above equation to find the value of t. It is equal to :
t = 3.551 seconds
or
t = 8.449 seconds
So, the tide of 6 ft is at 3.551 seconds and 8.449 seconds. Hence, this is the required solution.
Formula for potential energy is V=mgh, where m is mass in KG, g is earth acceleration (10 m/s^2), and h its height in meters. We know mass, acceleration is constant and also known, we know height also. Lets substitute
V=75*10*300=225000[J]=225[kJ] - its the answer
Answer:
It can be seen from the operation of pin-hole camera, formation of shadows and eclipse.
Explanation:
The phenomenon of light traveling in a straight line is known as rectilinear propagation of light.
One this evidence can be seen from the operation of pin-hole camera, which depends on rectilinear propagation of light
Also two natural effects that result from the rectilinear propagation of light are the formation of Shadows and Eclipse.