Answer:
L = 2.8 cm
Explanation:
Period T = 4 / 12 = 1/3 s
T = 2π√(L/g)
L = (T/2π)²g
L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm
Answer:
C.) 1.5 kg
Explanation:
Start with the equation:

Plug in what you know, and solve:

Find matching soluation:
C.) 1.5 kg
<span>The relationship between wavelength, frequency and energy of Electromagnetic Radiation is given by
E = hf = hc/lamba -------(1)
So from (1) there's a linear relationship between E and f. The higher the frequency, f, the higher the energy E.
Also from (1) it is obvious that the lower the wavelength, lambda, the higher the energy, E.
This means the answer is D.</span>
The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
<h3>How to determine the friction factor</h3>
Using the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 × 
friction factor = 0. 52 × 
friction factor = 0. 52 × 0. 55
friction factor 
b. When V = 3mls
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 0. 185
Friction factor 
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 ×
×
× 
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss =
×
×
× 
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
Learn more about friction here:
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The answer is "156.6 m/s".
This is how we calculate this;
-N + mg = ma = mv²/r
For "weightlessness" N = 0, so
0 = mg - mv²/r
g - v²/r = 0
v =√( gr)
g = 9.8 and r = 2.5km = 2500 m
v = √(9.8 x 2500)
= 156.6 m/s