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°.
Answer: Things continue doing what they are doing unless a force is applied to it. Objects have a natural tendency to resist change. This is INERTIA. Heavier objects (objects with more mass) are more difficult to move and stop. Heavier objects (greater mass) resist change more than lighter objects, so true
Explanation:
Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop. The Cadillac has more of a tendency to stay stationary (or continue moving), and resist a change in motion than a bicycle.
Answer:
the final kinetic energy is 0.9eV
Explanation:
To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:
you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is
![E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV](https://tex.z-dn.net/?f=E_%7Bn_2-n_1%7D%3D-13.6eV%5B%5Cfrac%7B1%7D%7Bn_2%5E2%7D-%5Cfrac%7B1%7D%7Bn_1%5E2%7D%5D%5C%5C%5C%5CE_%7B2-1%7D%3D-13.6eV%5B%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B1%7D%5D%3D-10.2eV)
-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:
Answer:
The frequency of the wheel is the number of revolutions per second:
f= \frac{N_{rev}}{t}= \frac{10}{1 s}=10 Hz
And now we can calculate the angular speed, which is given by:
\omega = 2 \pi f=2 \pi (10 Hz)=62.8 rad/s in the clockwise direction.
Explanation:
