frequency=4Hz
wavelength=5m
amplitude=1/2×2=1m
period=1/frequency
1/4=0.25seconds.
velocity=wavelength×frequency
=5×4
=20m/s
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.
Learn more about distance here: brainly.com/question/23848540
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The centripetal acceleration of an object is given by the relation,
where Ac = centripetal acceleration = 
R = radius of rotation = 15 m
V = speed of astronaut
Hence, 
solving this we get, V = 38.34 m/s
Answer:
Explanation:
Let T be the tension in the cord.
Impulse by cord = change in momentum of block A .
T x 5s = 10 ( 2 -0) = 20
T = 4 poundal .
acceleration of block B = 2 / 5 = 0.4 m /s²
Net force applied on A = m ( g + a ) where m is mass of block B , a is acceleration of block B .
= 8 ( 32 + .4 ) = 259.2 poundal
Frictional force on block A = 259.2 - 4 = 255.2 poundal
μ x 10 x 32 = 255.2
320μ = 255.2
μ =0 .8 .
