Answer:
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Answer:
33.6 m
Explanation:
Given:
v₀ = 0 m/s
a = 47.41 m/s²
t = 1.19 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (1.19 s) + ½ (47.41 m/s²) (1.19 s)²
Δx = 33.6 m
Answer:
I think the answers March 21
Answer:
Explanation:
a ) angular frequency ω = 
k is spring constant and m is mass attached
ω = 
= 3.6515 rad / s
frequency of oscillation n = 3.6515 / (2 x 3.14)
= .5814 s⁻¹
x = .1 mcos(ωt)
= .1 mcos(3.6515t)
b ) maximum speed = ωA , A is amplitude
= 3.6515 x .1
= .36515 m /s
36.515 cm /s
maximum acceleration = ω²A
= 3.6515² x .1
= 1.333 m / s²
c ) Kinetic energy at displacement x
= 1/2 m ω²( A²-x²)
potential energy =1/2 m ω²x²
so 1/2 m ω²( A²-x²) = 1/2 m ω²x²
A²-x² = x²
2x² = A²
x = A / √2
We use 1/o + 1/i = 1/f where o is the distance of the object, i as distance of the image and f is the focal length.
Substituting, <span>1/ 100 + 1 / i = - 1 /25 </span>
<span>i = - 20 cm </span>
<span>For the case of the problem,</span>
<span>o = (20 + 30) = 50 cm </span>
<span>f = 33.33. </span>Using 1<span> / i + 1 / o = 1/f , </span><span> </span><span>i = 100 cm </span>
<span>M = magnification = - i / o </span>
<span>m1 = -(-20)/100 = 20/100 = 0.2 </span>
<span>m2 = -100/50 = -2 </span>
<span>M = m1*m2 = -2 x 0.2 = -0.4.</span>
