A wire has a current density of 6.25 × 10 6 A / m 2 6.25×106 A/m2 . If the cross-sectional area of the wire is 1.79 mm 2 1.79 mm
1 answer:
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Answer:
Explanation:
I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.
The relation between the linear acceleration and the angular acceleration is
a = R α .... (1)
According to the diagram,
T x R = I x α
T x R = I x a / R from equation (1)
T = I x a / R² .... (2)
mg - T = ma .... (3)
Substitute the value of T from equation (2) in equation (3)
T is the acceleration in the system
Substitute the value of a in equation (2)
This is the tension in the string.
Answer:
See the answers below
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
<u>First case</u>
Vf = 6 [m/s]
Vo = 2 [m/s]
t = 2 [s]
<u>Second case</u>
Vf = 25 [m/s]
Vo = 5 [m/s]
a = 2 [m/s²]
<u>Third case</u>
Vo =4 [m/s]
a = 10 [m/s²]
t = 2 [s]
<u>Fourth Case</u>
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
<u>Fifth case</u>
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]