molecules of water are never destroyed - they go through various uses in a cycle of re-use. beginning in the ocean. a water molecue is attached to the wet suit of a deep sea diver. when the diver gets back on his boat, the water molecule leaves the ocean. Diver dry his suit under the sun. The water molecule is evaporated to the air. It meets up with more water molecules to form cloud. Cloud becomes rain over ground. Rain drains into stream which merges into river. River runs out to the ocean and the water cycle starts anew.
Answer
Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)
Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.
ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)
Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC
30.7pC/εo = 3.47 V∙m <----- C)
ic(0.5ns) = 29.7ma <----- D)
I need someone to help me with my finals!!!!
Answer:
a.) F = 3515 N
b.) F = 140600 N
Explanation: given that the
Mass M = 74kg
Initial velocity U = 7.6 m/s
Time t = 0.16 s
Force F = change in momentum ÷ time
F = (74×7.6)/0.16
F = 3515 N
b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds
Change in momentum = 74×7.6 + 74×7.6
Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s
F = 1124.8/0.0080 = 140600 N
Answer:
a) 4.2m/s
b) 5.0m/s
Explanation:
This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.
The problem is also an illustration of elastic collision where there is no loss in kinetic energy.
Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses 
and 
whose respective velocities before collision are 
and 
;
where 
and 
are their respective velocities after collision.
Given;
Note that =0 because the second mass was at rest before the collision.
Also, since the two masses are equal, we can say that 
so that equation (1) is reduced as follows;
m cancels out of both sides of equation (2), and we obtain the following;
a) When 
, we obtain the following by equation(3)
b) As stops moving 
, therefore,
