A car is traveling due north at 23.6 m>s.
Find the velocity of the car after 7.10 s if its
acceleration is
The acceleration is known to be: a(t) = 1.7 m/s2.
We must integrate over time to obtain the velocity function, and the results are:
v(t) = (1.7m/s^2)
*t + v0
If we suppose that we begin at 23.6 m/s, then the initial velocity is: v0 = 23.6 m/s, where v0 is the beginning velocity.
The velocity formula is then: v(t) = (1.7m/s2).
*t + 23.6 m/s
We now seek to determine the value of t such that v(t) = 27.8 m/s.
Consequently, v(t) = 27.8 m/s = (1.7 m/s2)
*t + 23.6 m/s = (1.7 m/s2) 27.8 m/s - 23.6 m/s
t = 2.5 seconds when *t 4.2 m/s = (1.7 m/s/2)
At such acceleration, 2.5 seconds are required.
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The alpha particle is emitted at 4235 m/s
Explanation:
We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:
where:
is the mass of the original nucleus
is the initial velocity of the nucleus
is the mass of the alpha particle
is the final velocity of the alpha particle
is the mass of the daughter nucleus
is the final velocity of the nucleus
Solving for , we find the final velocity of the alpha particle:
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C. Is the correct answer for this question
Answer:
Explanation:
Work done by the person depends on its mass and the height raised. It also depends on the acceleration due to gravity.
As the height raised and the mass of person is same. The value of acceleration due to gravity is also constant So, the work done is also same in both the cases.
a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
