A force of 880 newtons stretches a spring 4 meters. A mass of 55 kilograms is attached to the end of the spring and is initially
released from the equilibrium position with an upward velocity of 12 m/s. Find the equation of motion.
1 answer:
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The magnitude of the electrostatic force between two charges is given by:
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We can see that the magnitude of the force is directly proportional to the charges. This means that when one of the charges is doubled, the magnitude of the electrostatic force will double as well, so the correct answer is
A) <span>The magnitude of the electrostatic force doubles</span>
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We can see that the magnitude of the force is directly proportional to the charges. This means that when one of the charges is doubled, the magnitude of the electrostatic force will double as well, so the correct answer is
A) <span>The magnitude of the electrostatic force doubles</span>
Answer:
(a) The angle of projection is 63 degree.
(b) The velocity of projection is 24.5 m/s.
Explanation:
Height, h = 1 m
horizontal distance, d = 50 m
time, t = 4.5 s
Let the initial velocity is u and the angle is A.
(a) Horizontal distance = horizontal velocity x time
50 = u cos A x 4.5
u cos A = 11.1 .....(1)
Use second equation of motion in vertical direction
Divide (2) by (1)
tan A = 1.97
A = 63 degree
(b) Substitute the value of A in equation (2)
u x sin 63 = 21.8
u = 24.5 m/s