Answer: Im not entirly sure but I think It's D all the above. I think all but B because I never really heard of that but if you look in our history I think that happen im not sure I would wait untill you know that somone knows for sure.
Explanation:
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
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so 
= A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Answer:
Explanation:
We shall apply law of conservation of momentum to know velocity after collision . Let it be v .
total momentum before collision = total momentum after collision
15 x 1.5 - 12 x .75 = ( 15 + 12 ) v
v = .5 m /s
kinetic energy before collision
1/2 x 15 x 1.5² + 1/2 x 12 x .75²
= 16.875 + 3.375
= 20.25 J
kinetic energy after collision
= 1/2 x ( 15 + 12 ) x .5²
= 3.375 J
Loss of energy = 16.875 J
This energy appear as heat and sound energy that is produced during collision .
