A wagon is pulled at an angle of 30 degrees to the horizontal.
Answer:
1456 N
Explanation:
Given that
Frequency of the piano, f = 27.5 Hz
Entire length of the string, l = 2 m
Mass of the piano, m = 400 g
Length of the vibrating section of the string, L = 1.9 m
Tension needed, T = ?
The formula for the tension is represented as
T = 4mL²f²/ l, where
T = tension
m = mass
L = length of vibrating part
F = frequency
l = length of the whole part
If we substitute and apply the values we have Fri. The question, we would have
T = (4 * 0.4 * 1.9² * 27.5²) / 2
T = 4368.1 / 2
T = 1456 N
Thus, we could conclude that the tension needed to tune the string properly is 1456 N
S=56, u=0, v=33, a=?, t=3.4
v=u+at
33=3.4 a
a = 9.7m/s^2
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