A tow rope pulls a 1450 kg truck, giving it an acceleration 1.25 m/s^2. What force does the rope exert?
1 answer:
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Answer:68.15m/s
Explanation:
<u><em>Given: </em></u>
v₁=15m/s
a=6.5m/s²
v₁=?
x=340m
<u><em>Formula:</em></u>
v₁²=v₁²+2a (x)
<u>Set up:</u>
=
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Answer:
(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz
(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz
Explanation:
(a) The fundamental, f₁, frequency is given as follows;
Where;
T = The tension in the string
μ = The linear density of the string
L = The length of the string
f₁ = The fundamental frequency = 560 Hz
If the tension in the string is increased by 15%, we will have;
Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz
(b) When the string length is decreased by one-third, we have;
The new length of the string, = 2/3·L
The value of the fundamental frequency will then be given as follows;
When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.