The distance traveled by the particle at the given time interval is 0.28 m.
<h3>
Position of the particle at time, t = 0</h3>
The position of the particle at the given time is calculated as follows;
x = 2 sin2(t)
y = 2 cos2(t)
x(0) = 2 sin2(0) = 0
y(0) = 2 cos2(0) = 2(1) = 2
<h3>
Position of the particle at time, t = 4</h3>
x = 2 sin2(t)
y = 2 cos2(t)
x(4) = 2 sin2(4) = 0.28
y(4) = 2 cos2(4) = 2(1) = 1.98
<h3>Distance traveled by the particle at the given time interval</h3>
d = √[(x₄ - x₀)² + (y₄ - y₀)²]
d = √[(0.28 - 0)² + (1.98 - 2)²]
d = 0.28 m
Thus, the distance traveled by the particle at the given time interval is 0.28 m.
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Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,
Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?
a = -1.47 m/ (a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/ (since same friction force is applied)
s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
<u>Answer</u>
3.7 Km south
<u>Explanation</u>
The definition of displacement is the distance traveled in a specific direction. It is the vector quantity. We add displacements like the way we add vectors.
Taking the direction towards North to be positive (+1.7 Km), the distance towards south would be negative (-5.4 Km).
Now lets add the two values.
(+1.7) + (-5.4) = 1.7 - 5.4
= - 3.7 Km But negative was towards south.
∴ Answer = 3.7 Km south.
Light can be seen as an electromagnetic wave.
What happens when two waves, with the same frequency, superpose is called interference.
If at a certain point two waves arrive both with a crest, we have constructive interference and the amplitudes sum up, reaching the maximum value, resulting in bright spots.
If at a certain point one of the waves arrives with a crest and the other wave arrives with a trough, we have destructive interference, and the two amplitudes cancel out, resulting in dark spots.
Therefore, t<span>he dark bands on the wall are from destructive interference.</span>
