PLEASE HELP ASAPExplain how can a ripple tank be used to view wave pattern.
1 answer:
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Answer:
the tension in the part of the cord attached to the textbook is 7.4989 N
Explanation:
Given the data in the question;
As illustrated in the image below;
first we determine the value of the acceleration,
along vertical direction; we use the second equation of motion;
y = ut + a
t²
we substitute;
0 m/s for u, 1.29 m for y, 0.850 s for t,
1.29 = 0×0.850 + ×a
×(0.850)²
1.29 = 0.36125a
a = 1.29 / 0.36125
a = 3.5709 m/s²
Now when the text book is moving with acceleration , the dynamic equation will be;
T₁ = m₁a
where m₁ is the mass of the text book ( 2.10 kg )
a is the vertical acceleration ( 3.5709 m/s² )
so we substitute
T₁ = 2.10 × 3.5709
T₁ = 7.4989 N
Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N
Answer:
Explanation:
mass of proton, m = 1.67 x 10^-27 kg
speed of proton, v = 350 km/s = 350,000 m/s
Momentum of proton, p = mass x speed
p = 1.67 x 10^-27 x 350000 = 5.845 x 10^-22 kg m /s
uncertainty in momentum, Δp = 0.1 % of p
Δp =
According to the principle
where, Δx be the uncertainty in position