IP Force to Hold a Baseball A person holds a 1.42-N baseball in his hand, a distance of 34.0 cm from the elbow joint, as shown i
2 answers:
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Answer:
a. 2.4 ×109 N ⋅ m2/s
b. 48.3 N⋅s /m2
c. 8.00×104W
Explanation:
See Attached file for explanation
Answer:
j
Explanation:
x = 4 t^2 - 2 t - 4.5
Position at t = 3 s
x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m
Velocity at t = 3 s
v = dx / dt = 8 t - 2
v ( t = 3 s) = 8 x 3 - 2 = 22 m/s
Acceleration at t = 3 s
a = dv / dt = 8
a ( t = 3 s ) = 8 m/s^2
When is the velocity = 0
v = 0
8 t - 2 = 0
t = 0.25 second
When is the position = 0
x = 0
4 t^2 - 2 t - 4.5 = 0
t = 1.4 second
Answer:
29.76245 rad/s², -117.80972 rad/s²
28.2743 rad/s
3.95833
Explanation:
= Final angular velocity
= Initial angular velocity
= Angular acceleration
= Angle of rotation
t = Time taken
Equation of rotational motion
Angular acceleration during speed up is 29.76245 rad/s²
Angular acceleration during spin down is -117.80972 rad/s²
Angular speed is given by
Maximum angular speed reached by the flywheel is 28.2743 rad/s
The ratio would be
Check the attached file for the answer.
