A motor spins upward the flywheel with a persistent torque
of 50N⋅m.
What time does it take the flywheel to get to the top speed?
From the equation:
Tj = J*dω/dt
you can get the two equations:
Δt1= J1*Δω/Tj = 240*125.66/50 = 603.17 sec
Δt2= J2*Δω/Tj = 120*125.66/50 = 301.58 sec
Answer:
I = 0.002593 A = 2.593 mA
Explanation:
Current density = J = (3.00 × 10⁸)r² = Br²
B = (3.00 × 10⁸) (for ease of calculations)
The current through outer section is given by
I = ∫ J dA
The elemental Area for the wire,
dA = 2πr dr
I = ∫ Br² (2πr dr)
I = ∫ 2Bπ r³ dr
I = 2Bπ ∫ r³ dr
I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]
I = (Bπ/2) [R⁴ - (0.9R)⁴]
I = (Bπ/2) [R⁴ - 0.6561R⁴]
I = (Bπ/2) (0.3439R⁴)
I = (Bπ) (0.17195R⁴)
Recall B = (3.00 × 10⁸)
R = 2.00 mm = 0.002 m
I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]
I = 0.0025929449 A = 0.002593 A = 2.593 mA
Hope this Helps!!!
Answer:
the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.
Explanation:
a) Kinetic energy of block = potential energy in spring
½ mv² = ½ kx²
Here m stands for combined mass (block + bullet),
which is just 1 kg. Spring constant k is unknown, but you can find it from given data:
k = 0.75 N / 0.25 cm
= 3 N/cm, or 300 N/m.
From the energy equation above, solve for v,
v = v √(k/m)
= 0.15 √(300/1)
= 2.598 m/s.
b) Momentum before impact = momentum after impact.
Since m = 1 kg,
v = 2.598 m/s,
p = 2.598 kg m/s.
This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,
u = 2.598 kg m/s / 8 × 10⁻³ kg
= 324.76 m/s.
Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.
Bert would be going 34mph due to the taill winds being 6mph