This is called the Phi Phenomenon.
This is an illusion of movement created when two or more adjacent lights blink on and off in quick succession; when two adjacent stationary lights blink on and off in quick succession; we perceive a single light moving back and forth between them. It is an optical illusion of perceiving a series of still images, when viewed in rapid succession, as continuous motion.
Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s
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Answer:
Explanation:
If we have a periodic wave we need to satisfy the following basic relationship:
From the last formula we see that the velocity is proportional fo the frequency.
For this case we have the following info given by the problem:
We know that the frequency is the reciprocal of the period so we have this formula:
And if we replace we got:
Now since we have the value for the wavelength we can find the velocity like this:
And if we convert this into cm/s we got:
Answer:
r = 1.61 x 10^{11} m
Explanation:
energy radiated (H) = 2.7 x 10^31 W
surface temperature (T) = 11,000 k
assuming ε = 1 and taking σ = 5.67 x 10^{-8} W/m^{2}.K^{4}
we can find the radius of the star from the equation below
H = A x ε x σ x T^{4}
where area (A) = 4 x π x r^{2} (assuming it is a sphere)
therefore the equation becomes
H = 4 x π x r^{2} x ε x σ x T^{4}
2.7 x 10^31 = 4 x π x r^{2} x 1 x 5.67 x 10^{-8} x (11,000)^{4}
r = 
r = 1.61 x 10^{11} m
