Answer:
In a coiled spring, the particles of the medium vibrate to and fro about their mean positions at an angle of
A. 0° to the direction of propagation of wave
Explanation:
The waveform of a coiled spring is a longitudinal wave, which is made up of vibrations of the spring which are in the same direction as the direction of the wave's advancement
As the coiled spring experiences a compression force and is then released, it experiences a sequential movement of the wave of the compression that extends the length of the coiled spring which is then followed by a stretched section of the coiled spring in a repeatedly such that the direction of vibration of particles of the coiled is parallel to direction of motion of the wave
From which we have that the angle between the direction of vibration of the particles of the coiled spring and the direction of propagation of the wave is 0°.
<span>The equation that relates force and torque is Torque = Force*length*sin(angle). Plug in the given values to the equation (convert length to meters first): -13 = Force * 0.35 * sin(30 degrees). Rearrange to solve for Force: Force = -13/(0.35 * 0.5) = -74.3 N (which is the same as a downward force of magnitude 74.3 N.</span>
Explanation:
Draw a free body diagram of the bucket. There are two forces:
Tension force T pulling up.
Weight force mg pulling down.
Sum of forces in the y direction:
∑F = ma
T − mg = ma
T = mg + ma
T = m (g + a)
Given m = 4.0 kg, g = 9.81 m/s², and a = 2.8 m/s²:
T = (4.0 kg) (9.81 m/s² + 2.8 m/s²)
T = 50.44 N