The force shown in the attached figure is the net eastward force acting on a ball. The force starts rising at t = 0.012 s, falls
back to zero at t = 0.062 s, and reaches a maximum force of 35 N at the peak. Determine with an error no bigger than 25% (high or low) the magnitude of the impulse (in N-s) delivered to the ball. Hint: Do not use J = FΔt. Look at the figure. Find the area of a nearly equally sized triangle.
1 answer:
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To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:
The angular displacement is given as the form:
In the equlibrium we have to and in the given position we have to
Derived the expression we will have the equivalent to angular velocity
Replacing,
Finally
Therefore the maximum angular displacement is 9.848°
Answer:
The answer is given below
Explanation:
u is the initial velocity, v is the final velocity. Given that:
a)
The final velocity of cart 1 after collision is given as:
The final velocity of cart 2 after collision is given as:
b) Using the law of conservation of energy:
Answer:
9.21954 m/s
54 m/s²
Angle is zero
Explanation:
r = Radius of arm = 1.5 m
= Angular velocity = 6 rad/s
The horizontal component of speed is given by
The vertical component of speed is given by
The resultant of the two components will give us the velocity of hammer with respect to the ground
The velocity of hammer relative to the ground is 9.21954 m/s
Acceleration in the vertical component is zero
Net acceleration is given by
Net acceleration is 54 m/s²
As the acceleration is towards the center the angle is zero.