The speed of sound is greater in ice (4000 m/s), then in water (1500 m/s), then in air (340 m/s). The explanation for this is the differente state of the matter in the three cases.
In fact, sound waves travel faster in solids (like ice), then in liquids (like water), then in gases (like air). This is because the speed of the sound wave depends on the density of the medium: the greater the density, the faster the sound wave. This can be easily understood by thinking at how a sound wave propagates: a sound wave is a vibration of molecules, which is transmitted throughout the medium by collision of the molecules. Therefore, the smaller the spacing between the molecules (such as in solids), the more efficient is the propagation, and so the sound wave is faster. On the contrary, there is a large spacing between molecules in gases (such as in the air), so there are less collisions between the molecules and so the wave is not transmitted efficiently, and so it has less velocity.
Explanation:
The given data is as follows.
C = 
R = 
ohm
C
Q = 
Formula to calculate the time is as follows.
![Q_{t} = Q_{o} [e^{\frac{-t}{\tau}]](https://tex.z-dn.net/?f=Q_%7Bt%7D%20%20%3D%20Q_%7Bo%7D%20%5Be%5E%7B%5Cfrac%7B-t%7D%7B%5Ctau%7D%5D)
![13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]](https://tex.z-dn.net/?f=13.5%20%5Ctimes%2010%5E%7B-6%7D%20%3D%20100%20%5Ctimes%2010%5E%7B-6%7D%20%5Be%5E%7B%5Cfrac%7B-t%7D%7B2%7D%7D%5D)
0.135 = 
= 7.407
t = 4.00 s
Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.
Answer:
According to the travellers, Alpha Centauri is <em>c) very slightly less than 4 light-years</em>
<em></em>
Explanation:
For a stationary observer, Alpha Centauri is 4 light-years away but for an observer who is travelling close to the speed of light, Alpha Centauri is <em>very slightly less than 4 light-years. </em>The following expression explains why:
v = d / t
where
- v is the speed of the spaceship
- d is the distance
- t is the time
Therefore,
d = v × t
d = (0.999 c)(4 light-years)
d = 3.996 light-years
This distance is<em> very slightly less than 4 light-years. </em>
Answer:
Atomic and molecular collision processes are the physical interactions of atoms and molecules when they are brought into close contact with each other and with electrons, protons, neutrons or ions. This includes energy-conserving elastic scattering and inelastic scattering.
The work done by a constant force in a rectilinear motion is given by:
where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.
In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:
Therefore, the total work done is 578.123 J and the answer is option E
