The equation for distance is d = st. If a car has a speed of 35 m/s, how long will it take to go 100 m? A.2.9 s B.3500 s C.65 s
2 answers:
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Answer:
n=6.56×10¹⁵Hz
Explanation:
Given Data
Mass=9.1×10⁻³¹ kg
Radius distance=5.3×10⁻¹¹m
Electric Force=8.2×10⁻⁸N
To find
Revolutions per second
Solution
Let F be the force of attraction
let n be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by
F=mω²r......................where ω=2π n
F=m4π²n²r...............eq(i)
as the values given where
Mass=9.1×10⁻³¹ kg
Radius distance=5.3×10⁻¹¹m
Electric Force=8.2×10⁻⁸N
we have to find n from eq(i)
n²=F/(m4π²r)
Answer:
Explanation:
Given that,
h(t) = -9.8t² / 2 + 125t + 500
for t > 0.
At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.
We want to find the maximum height reached by rocket
Using mathematics maxima and minima
let find the turning point when dh/dt = 0
dh/dt = -9.8t + 125
dh / dt = 0 = -9.8t + 125
9.8t = 125
t = 125 / 9.8
t = 12.76s
Let find the turning point to know if this time t = 12.76 is maximum or minimum point
Let find d²h / dt²
d²h / dt² = -9.8
Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.
Then, the maximum height reached is
h = -9.8t² / 2 + 125t + 500
h = -9.8(12.76)² / 2 + 125(12.76) + 500
h = -797.80 + 1595 + 500
h = 1297.2 m
The maximum height reached is 1297.2 m
From the attachment, the maximum height is 1297.2m at t = 12.76sec.
Comment, the result are the same for both the analysis aspect and the graphical aspect.
