Answer:
a) The duration, during which the block remain in contact with the spring is 0.29 s
b) The period of the simple harmonic oscillatory motion depends only on the mass and spring constant, therefore when the speed is doubled, the duration of contact remains the same as 0.29 s.
Explanation:
Mass of the block = 465 g
Surface speed = 0.35 m/s
Spring constant , k = 54 N/m
= 0.58 s
a) Since the period for the oscillatory motion is 0.58 s, then the time when the block and spring remain in contact is T/2 = 0.29 s
b) When the speed is doubled, we have
Therefore, since T is only dependent on the mass, m and the spring constant, K, then the time it takes when the speed is doubled remain as
T /2 = 0.29 s
Answer:
0.2 T
Explanation:
Magnetic field is inversely proportional to the distance from wire since the distance is halved therefore magnetic field will be doubled.
Answer:
Q = 47.06 degrees
Explanation:
Given:
- The transmitted intensity I = 0.464 I_o
- Incident Intensity I = I_o
Find:
What angle should the principle axis make with respect to the incident polarization
Solution:
- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:
I = I_o * cos^2 (Q)
- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:
Q = cos ^-1 (sqrt (I / I_o))
- Plug the values in:
Q = cos^-1 ( sqrt (0.464))
Q = cos^-1 (0.6811754546)
Q = 47.06 degrees
