The stiffness constant of the spring is 68,290.3 N/m
<h3>
Stiffness constant of the spring</h3>
Apply the principle of conservation of energy;
U = K.E
¹/₂kx² = ¹/₂mv²
kx² = mv²
k = mv²/x²
where;
- v is speed = 60 km/h = 16.67 m/s
- x is the distance
k = (1300 x 16.67²)/(2.3²)
k = 68,290.3 N/m
Thus, the stiffness constant of the spring is 68,290.3 N/m.
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The wavelength of the standing wave at fourth harmonic is; λ = 0.985 m and the frequency of the wave at the calculated wavelength is; f = 36.84 Hz
Given Conditions:
mass of string; m = 0.0133 kg
Force on the string; F = 8.89 N
Length of string; L = 1.97 m
1. To find the wavelength at the fourth normal node.
At the fourth harmonic, there will be 2 nodes.
Thus, the wavelength will be;
λ = L/2
λ = 1.97/2
λ = 0.985 m
2. To find the velocity of the wave from the formula;
v = √(F/(m/L)
Plugging in the relevant values gives;
v = √(8.89/(0.0133/1.97)
v = 36.2876 m/s
Now, formula for frequency here is;
f = v/λ
f = 36.2876/0.985
f = 36.84 Hz
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In order to make things easier to describe and explain, let's call
the resistance of each bulb 'R', and the battery voltage 'V'.
a). In series, the total resistance is 3R.
In parallel, the total resistance is R/3.
Changing from series to parallel, the total resistance of the circuit
decreases to 1/9 of its original value.
b). In series, the total current is V / (3R) .
In parallel, the total current is 3V / R .
Changing from series to parallel, the total current in the circuit
increases to 9 times its original value.
c). In series, the power dissipated by the circuit is
(V) · V/3R = V² / 3R .
In parallel, the power dissipated by the circuit is
(V) · 3V/R = 3V² / R .
Changing from series to parallel, the power dissipated by
the circuit (also the power delivered by the battery) increases
to 9 times its original value.
The answer to this is C jellyfish i am positive
hope this helps
