An ideal monatomic gas expands isobarically, doubling the volume, from a state A to state B. It is then compressed isothermally
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Answer:
The speed of the bob when it passes the lowest point
Explanation:
Given data
When a pendulum swinging back & forth then at highest point the velocity is zero and lowest point velocity is maximum.
Velocity at lowest point is given by
Therefore the speed of the bob when it passes the lowest point
The magnetic force acting on a charged particle moving perpendicular to the field is:
= qvB
is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.
The centripetal force acting on a particle moving in a circular path is:
= mv²/r
is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.
If the magnetic force is acting as the centripetal force, set equal to
and solve for v:
qvB = mv²/r
v = qBr/m
Due to the work-energy theorem, the work done on the proton by the potential difference V becomes the proton's kinetic energy:
W = KE
W is work, KE is kinetic energy
W = Vq
KE = 0.5mv²
Therefore:
Vq = 0.5mv²
Substitute v = qBr/m and solve for V:
V = 0.5qB²r²/m
Given values:
m = 1.67×10⁻²⁷kg (proton mass)
B = 0.750T
q = 1.60×10⁻¹⁹C (proton charge)
r = 1.84×10⁻²m
Plug in the values and solve for V:
V = (0.5)(1.60×10⁻¹⁹)(0.750)²(1.84×10⁻²)²/1.67×10⁻²⁷
V = 9120V