There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.
You need to figure out t4 to know the tension in the string.
Since the whole thing is not moving t1 + t2 + t3 = t4.
torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)
t1 =3.2 * 44g
t2 = 7 * 49g
t3 = 3.5 * 24g
t4 = t1 + t2 + t3 = 5570,118
The t4 also is given by:
t4 = r * T * sin Ф
r = 7
Ф = 32°
T: tension in the string
T = t4 / (r * sinФ)
T = t4 / (7 * sin(32°))
T = 1501,6 N
Answer:
The frequency of the oscillation is 0.9Hz
Explanation:
This problem bothers on simple harmonic motion of a spring
Given data
Mass of the child m= 25kg
Spring constant k=791 N/m
Amplitude a= 31cm
But the period of the motion as a result of the adults sholve is expressed as
T=2π√m/k
T=2*3.142√25/791
T=6.284√0.031
T=6.284*0.176
T=1.11 sec
But frequency F=1/T
F=1/1.11
F=0.9Hz
Answer:shorter higher
Explanation:Compare to visible light, the wavelength of X-rays is shorter and then frequency is higher
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