A copper wire of cross sectional area 2.0mm2 carries a current of 10A. How many electrons pass through a given cross section of
2 answers:
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Explanation:
Given that,
Length of the spring, l = 50 cm = 0.5 m
Mass connected to the end, m = 330 g = 0.33 kg
The mass is released and falls, stretching the spring by 30 cm before coming to rest at its lowest point. On applying Newton's second law, 10 cm below the release point, x = 15 cm
(a) When the mass is connected, the force of gravity is balanced by the force in spring.
(b) The amplitude of the oscillation will be 15 cm as it is half of the total distance travelled.
(c) The frequency of the oscillation is given by :
Hence, this is the required solution.
Answer:
d = 19.796m
Explanation:
Since the ball is in the air for 4.02 seconds, the ball should reach the maximum point from the ground in half the total time, therefore, t=2.01s to reach maximum height. At the maximum height, the velocity in the y-direction is 0.
So we know t=2.01, vi=0, g=a=9.8m/s and we are solving for d.
Next, you look for a kinematic equation that has these parameters and the one you should choose is:
Now by substituting values in, we get
d = 19.796m