A forklift lifts 5 boxes from the ground to a height of 2 meters (m). The boxes push down with a force of 1000 newtons (N). How
1 answer:
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Complete question:
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find
(a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(a) the force constant of the spring = 47 N/m
(b) the amplitude of the motion = 0.292 m
Explanation:
Given;
mass of the spring, m = 200g = 0.2 kg
period of oscillation, T = 0.410 s
total mechanical energy of the spring, E = 2 J
The angular speed is calculated as follows;
(a) the force constant of the spring
(b) the amplitude of the motion
E = ¹/₂kA²
2E = kA²
A² = 2E/k
Answer:
The magnitude of the magnetic field is .
Explanation:
Given that,
Charge,
Speed of the charged particle,
The angle between the velocity of the charge and the field is 56°.
The magnitude of force,
We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :
B is the magnetic field.
So, the magnitude of the magnetic field is . Hence, this is the required solution.