The force on a charged particle in a magnetic field is given by
the speed of the charged particle = 10842 m/s.
Explanation:
F= q V B sinθ
F=force=3.5 x 10⁻²N
q= charge= 8.4 x 10⁻⁴ C
B= magnetic field= 6.7 x 10⁻³ T
θ=35⁰
Thus the velocity is given by V=
V=(3.5 x 10⁻²)/[(8.4 x 10⁻⁴)(6.7 x 10⁻³)(sin35)]
V=10842 m/s
Answer:
v₁ = 37.5 cm / s
Explanation:
For this exercise we can use that angular and linear velocity are related
v = w r
in the case of the spool the angular velocity for the whole system is constant,
They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,
w = v₀ /r₀
for the outside of the spool r₁ = 1.5 cm
w = v₁ / r₁1
since the angular velocity is the same we set the two expressions equal
v1 =
let's calculate
v₁ =
v₁ = 37.5 cm / s
Answer:
The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp
Explanation:
<u>Polar moment of Inertia</u>
= 0.14374 in 4
<u>Maximum sustainable torque on the solid circular shaft</u>
=
= 3658.836 lb.in
= lb.ft
= 304.9 lb.ft
<u>Maximum sustainable torque on the tubular shaft</u>
=
= 3334.8 lb.in
= lb.ft
= 277.9 lb.ft
<u>Maximum sustainable power in the solid circular shaft</u>
=
= 4023.061 lb. ft/s
= hp
= 7.315 hp
<u>Maximum sustainable power in the tubular shaft</u>
=
= 3666.804 lb.ft /s
= hp
= 6.667 hp