Answer:
Frquency=3,994Hz
Explanation:
Tension =967N
Density of string (μ)=0.023g/cm
Length of the stretched spring=308cm
Fundamental frequency for nth harmonic :
Fn=n/2L(√T/μ)
Substituting the given values to find the frequency :
f1=1/2(308cm) *(0.01m/1cm)[(√967N)/(0.023g/cm)(0.1kg)/(0.1kg/m)/(1g/cm)]
=6.16m[(√967N)/0.0023kg/m)]
=3,994.20Hz
Approximately,
The frequency will be =3,994Hz
Becoming Cold Blooded im pretty sure
Answer:
<em>Air pockets.</em>
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Explanation:
Air pockets in the cooling system are bubbles of air trapped within the lines (hoses and pipes) of the cooling system. This air bubbles enter the cooling system usually during the process of filling the radiator coolant fluid (usually water), or replacing the water pump or the radiator hose during repairs or servicing of the cooling system. <em>The trapped air prevent pressure movement that is needed by the coolant to move the heat generated from the engine cylinder, resulting in heat build up</em>. The solution is to "bleed" the engine through the radiator lid or some air release valves.
I am not sure of the answer but I think that the answer is a
To solve this problem we will apply the concepts related to the Doppler effect. This is understood as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically this is given as,
Here,
v = Speed of the waves in the middle
= Speed of the receiver in relation to the medium (Positive if the receiver is moving towards the transmitter or vice versa)
= Speed of the source with respect to the medium (Positive if the source moves away from the receiver or vice versa)
Our values are given as,
Replacing,
Solving for the velocity of the source,
Therefore the speed of the other train is 26.1m/s
