B explanation : they are both filled to the same pint
Hooke's Law states that the extension is directly proportional to the force applied so:
F/x = constant
F₁/x₁ = F₂/x₂
2 / 0.02 = 1600 / x₂
x₂ = 16 m
Elastic work = 1/2 Fx
= 1/2 * 1600 * 16
= 12.8 kJ
Answer:
The magnetic field will be
, '2d' being the distance the wires.
Explanation:
From Biot-Savart's law, the magnetic field (
) at a distance '
' due to a current carrying conductor carrying current '
' is given by

where '
' is an elemental length along the direction of the current flow through the conductor.
Using this law, the magnetic field due to straight current carrying conductor having current '
', at a distance '
' is given by

According to the figure if '
' be the current carried by the top wire, '
' be the current carried by the bottom wire and '
' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the
symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by
symbol.
Given
and 
Therefore, the magnetic field (
) at 'P' due to the top wire

and the magnetic field (
) at 'P' due to the bottom wire

Therefore taking the value of
the net magnetic field (
) at the midway between the wires will be

Any fracture or system of fractures along which Earth moves is known as a fault.
Answer: b. fault.
A. because as the merry-go-round spins the child accelerates towards the center of the merry-go-round at a uniform rate.