To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion
From the definition we know that the frequency can be expressed as

Where,


Therefore the frequency would be given as


The frequency is directly proportional to the angular velocity therefore



Now the maximum speed from the simple harmonic movement is given by

Where
A = Amplitude
Then replacing,


Therefore the maximum speed of a point on the string is 3.59m/s
To change from mass to weight is Fw = 30 kg * 9.8 m/s^2 = 294 N. To change from weight to mass divide by gravity (9.8 m/s^2).
Answer:
See below
Explanation:
See attached diagram
280 km east then 190 km north
Use Pythagorean theorem to find the resultant displacement
d^2 = 280^2 + 190^2
d = 338.4 km
Angle will be arctan ( 190/280) = 34.16 °
Answer:
d
a balanced force acted on it and propelled it to 4,000 km/hr
Explanation:
For the neutrons star which is moving in outer space at 4,000 km/hr, it could only be possible as a result of the balanced force which had already acted on it. <em>This is based on newton's law of motion which states that 'To every action, there is equal and opposite reaction'. </em>
Answer:
Explanation:
T = 2π
(T / 2π)² = L/g
g = 4π²L/T²
g = 4π²(0.75000)/(1.7357)²
g = 9.82814766...
g = 9.8281 m/s²