Under which condition would time periods of daylight and darkness be equal everywhere on Earth all year? A. if Earth revolved ar
2 answers:
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Answer:
distance between the dime and the mirror, u = 0.30 m
Given:
Radius of curvature, r = 0.40 m
magnification, m = - 2 (since,inverted image)
Solution:
Focal length is half the radius of curvature, f =
f =
Now,
m = -
- 2 = -
= 2 (2)
Now, by lens maker formula:
v = (3)
From eqn (2):
v = 2u
put v = 2u in eqn (3):
2u =
2 =
2(u - 0.20) = 0.20
u = 0.30 m
- initial velocity=u=24m/s
- Acceleration=a=4m/s^2
- Distance=s=96m
- Final velocity=v
Using 3rd equation of kinematics
Answer:
ω = 1.83 rad/s clockwise
Explanation:
We are given:
I1 = 3.0kg.m2
ω1 = -5.4rad/s (clockwise being negative)
I2 = 1.3kg.m2
ω2 = 6.4rad/s (counterclockwise being positive)
By conservation of the momentum:
I1 * ω1 + I2 * ω2 = (I1 + I2) * ω
Solving for ω:
Since it is negative, the direction is clockwise.