Answer:
first one is b 2nd one is a 3rd is c and the 4th one is c also
Explanation: have a nice day
Answer:
To create a second harmonic the rope must vibrate at the frequency of 3 Hz
Explanation:
First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,
f₁ = v/2L
where,
v = speed of wave = 36 m/s
L = Length of rope = 12 m
f₁ = fundamental frequency
Therefore,
f₁ = (36 m/s)/2(12 m)
f₁ = 1.5 Hz
Now the frequency of nth harmonic is given in general, as:
fn = nf₁
where,
fn = frequency of nth harmonic
n = No. of Harmonic = 2
f₁ = fundamental frequency = 1.5 Hz
Therefore,
f₂ = (2)(1.5 Hz)
<u>f₂ = 3 Hz</u>
You are given the mass of a sphere that is 26 kg sphere and it is released from rest when θ = 0°. You are also given the force of the spring that is F = 100 N. You are asked to find the tension of the spring. Imagine that the sphere is connected to a spring. The spring exerts a tension and the spring exerts gravitational pull. This will follow the second law of newton.
T - F = ma
T = ma + F
T = 26kg (9.81m/s²) + 100 N
T = 355.06 N
Newtons (N) measure force
The spring constant is 147 N/m
Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm
We need to find the spring constant
A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.
The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring
We know that F = kx
300(9.8)= k (0.02)
k = 147.15 N/m
Rounding off to the nearest is 147N/m
The spring constant is 147N/m
Learn more about Hooke's law here
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