In zero order reactions the rate of reaction is independent of reactant concentrations. That is the rate of the reaction does not vary with increasing nor decreasing reactants concentrations. On the other hand, first order reaction are reactions in which the rate of reaction is directly proportional to the concentration of the reacting substance (reactants). In this case, i believe the rate of reaction will triple( increase by a factor of 3)
Answer:
A. it's the only answer that makes sense. if I'm wrong sorry
Answer:
20 m
Explanation:
From the equation of motion,
S = ut+1/2gt²................................. Equation 1
Where S = Height, u = initial velocity, t = time, g = acceleration due to gravity.
Note: Because the rocked is being dropped from a height, acceleration due to gravity is positive (g), and initial velocity (u) is negative
Given: t = 2.0 s, g = 10 m/s², u = 0 m/s (dropped from height)
Substituting into equation 1
S = 0(2) + 1/2(10)(2)²
S = 5(4)
S = 20 m
Hence the height of the the cliff above the pool is 20 m
Answer:
Explanation:
(b) The initial velocity is added to that due to acceleration by gravity. The velocity is increased linearly by gravity at the rate of 9.8 m/s². The average velocity of the pebble will be its velocity halfway through the 2-second time period.* That is, it will be ...
4 m/s + (9.8 m/s²)(2 s)/2 = 13.8 m/s . . . . average velocity
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(a) The distance covered in 2 seconds at an average velocity of 13.8 m/s is ...
d = vt
d = (13.8 m/s)(2 s) = 27.6 m
The water is about 27.6 m below ground.
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* We have chosen to make use of the fact that the velocity curve is linear, so the average velocity is half the sum of initial and final velocities:
vAvg = (vInit + vFinal)/2 = (vInit + (vInit +at))/2 = vInit +at/2
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If you work this in a straightforward way, you would find distance as the integral of velocity, then find average velocity from the distance and time.
