The force exerted by a magnetic field on a wire carrying current is:

where I is the current, L the length of the wire, B the magnetic field intensity, and

the angle between the wire and the direction of B.
In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is

, so we can re-arrange the formula and substitute the numbers to find B:
The energy becomes 0.50 times in 6.72 s.
Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.
Et=0.50E
1
k(4₂)² = (0.5) - kA²
(4₂)² = (0.5) A²
At = 0.71A
So, the amplitude of the oscillator becomes 0.71 times its initial ar
0.71A = = A(0.96)¹2
log(0.71)
log(0.96)
8.4
n=
So, the time taken for n oscillation is obtained as,
t = n (0.800 s)
= (8.4) (0.800)
= 6.72 s
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Mantle convection<span> is the slow creeping motion of Earth's solid silicate </span>mantle<span> caused by </span>convection<span> currents carrying heat from the interior of the Earth to the surface.
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Kinetic of automobile
Mass m = 1,250 Kg; V = 11 m/s
Formula: K.E = 1/2 mV²
K.E = 1/2(1,250 Kg)(11 m/s)²
K.E = 75,625 J
Speed required for insect to have the same kinetic energy as automobile
Mass of insect = 0.72 g convert to Kg m = 7.2 x 10⁻⁴ Kg
K.E = 1/2 mV² Derive V =?
V = 2 K.E/m
V = √2(75,625 J)/7.2 x 10⁻4 Kg
V = √2.1 x 10⁸ m²/s²
V = 14,491.34 m/s (velocity of insect)