To use Faraday's law for this problem, you will need to construct a suitable loop, around which you will integrate the electric
1 answer:
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Answer:
= 1647.92 N
Explanation:
Given:
length of the arm, r = 0.465 m
distance of forearm from elbow, r' = 2.15 cm = 0.0215 m
Mass of the forearm, M = 2.45 kg
Mass of the object, m = 6.55 kg
Let, the Force by bicep be,
under the motionless condition, the net moment about elbow is zero
thus,
× 0.0215 - Mg × (0.465/2) - mg × (0.465) = 0
or
× 0.0215 - 2.45 × 9.8 × (0.465/2) - 6.55 × 9.8 × (0.465) = 0
or
× 0.0215 - 5.582 - 29.848 = 0
or
= 1647.92 N
hence, the force exerted by the elbow is 1647.92 N
Answer:
speed of a particle = 2.31 × m/s
energy of proton required = 27.77 KeV
energy of electron required = 15.171 eV
Explanation:
given data
magnetic field of magnitude = 0.22 T
electric field of magnitude = 0.51 MV/m
to find out
speed of a particle and energy must protons have to pass through undeflected and energy must electrons have to pass through undeflected
solution
we know that force due to magnetic and electric field is express as
force due to magnetic field B = qvB ..............1
and force due to electric field E = qE .....................2
so without deflection force due to magnetic field = force due to electric field
so here qvB = qE
and V = ...................3
put here value
V =
speed of a particle = 2.31 × m/s
and
now energy of proton required will be here as
energy of proton required = mass of proton ×
put here value
energy of proton required = 1.67 × ×
energy of proton required = 4.45 × J
energy of proton required = 4.45 × J ÷ (1.602 ×
energy of proton required = 27777.777 eV
energy of proton required = 27.77 KeV
and
now we get here energy of electron required that is
energy of electron required = mass of electron ×
put here value
energy of electron required = 9.11 × ×
energy of electron required = 24.305× J
energy of electron required = 24.305 × J ÷ (1.602 ×
energy of electron required = 15.171 eV