Answer:
Therefore,
The frequency heard by the engineer on train 1

Explanation:
Given:
Two trains on separate tracks move toward each other
For Train 1 Velocity of the observer,

For Train 2 Velocity of the Source,

Frequency of Source,

To Find:
Frequency of Observer,
(frequency heard by the engineer on train 1)
Solution:
Here we can use the Doppler effect equation to calculate both the velocity of the source
and observer
, the original frequency of the sound waves
and the observed frequency of the sound waves
,
The Equation is

Where,
v = velocity of sound in air = 343 m/s
Substituting the values we get

Therefore,
The frequency heard by the engineer on train 1

Answer:
0.25m
Explanation:
Given parameters:
Spring constant , K = 120N/m
Work done = 3.75J
Unknown:
magnitude of extension = ?
Solution:
To solve this problem;
Work done =
kx²
K is the spring constant
x is the extension
3.75 =
x 120x²
3.75 = 60x²
x² = 0.06
x = √0.06 = 0.25m
Draw something similar to which has more mass in density. For example, a bowling ball and a volley ball. Both substances are similar in size but the bowling. All feels heavier than the volley ball. Label the volleyball for less density and the bowling ball with more density.