Answer:
The electric current in the wire is 0.8 A
Explanation:
We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

B= Magnetic field due to a straight and long wire that carries current
u= Free space permeability
I= Electrical current passing through the wire
a = Perpendicular distance from the wire to the point where the magnetic field is located
Magnetic Field Calculation
We cleared (I) of the formula (1):
Formula(2)

a =8cm=0.08m

We replace the known information in the formula (2)

I=0.8 A
Answer: The electric current in the wire is 0.8 A
Answer
given,
x = (3.9 cm)sin[(9.3 rad/s)πt]
general equation of displacement
x = A sin ω t
A is amplitude
now on comparing
c) Amplitude =3.9 cm
a) frequency =


f = 4.65 Hz
b) period of motion


T = 0.215 s
d) time when displacement is equal to x= 2.6 cm
x = (3.9 cm)sin[(9.3 rad/s)πt]
2.6 = (3.9 cm)sin[(9.3 rad/s)πt]
sin[(9.3 rad/s)πt] = 0.667
9.3 π t = 0.73
t = 0.025 s
Reverse faults is the most best answer
Answer:
Explanation:
Magnetic field at a a point R distance away
B = μ₀ / 4π X 2I / R where I is current
Magnetic field due to one current
= 10⁻⁷ x 2 x 24 / 1 x 10⁻³
48 x 10⁻⁴ T
Magnetic field due to other current
= 10⁻⁷ x 2 x 24 / 3x 10⁻³
16 x 10⁻⁴ T
Total magnetic field , as they act in opposite direction, is
= (48 - 16 ) x 10⁻⁴
32 x 10⁻⁴ T .
According to the plot, static friction force has a maximum magnitude of around 3.0 N, and kinetic friction has a magnitude of about 1.5 N.
The plot appears to be telling you the force required to get the yellow block moving along the table. If one applies less than 3.0 N of force, the block remains motionless. But as soon as it starts to slide, one need only apply 1.5 N of force to keep it moving (presumably at a constant speed).