acceleration times time falling
25*2
50 m*s^-1
Answer:
The brilliant galaxy centers are known as quasars. According to the astronomers quasars exists at the center o galaxies and in the center of galaxies there exists the black holes. Quasars are known as the brightest objects of the universe and they appear like a star in the telescopes.
Explanation:
The tiger's position at time <em>t</em> is given by
<em>x</em> (horizontal) = (4.5 m/s) <em>t</em>
<em>y</em> (vertical) = 7.5 m - 1/2 <em>gt</em> ²
Solve <em>y</em> = 0 for <em>t</em> to find the time it takes for the tiger to reach the ground :
0 = 7.5 m - 1/2 (9.8 m/s²) <em>t</em> ²
===> <em>t</em> = √(2 (7.5 m) / (9.8 m/s²)) ≈ 1.2 s
Evaluate <em>x</em> at this time :
<em>x</em> = (4.5 m/s) (1.2 s) ≈ 5.6 m
The original frequency of horn of Car A is 1071 Hz.
Explanation:
Doppler effect describes the change in the frequency of sound waves with respect to the observer. As the sound waves emitted from a source need to travel the air medium to reach observer, it will undergo loss in energy. So there will be change in its frequency compared to original frequency. Depending upon the direction of travel of source and observer the shifting of frequency will vary.

Here vo is the observer velocity and vs is the velocity of the source. So Vo = 15 m/s as car B is the observer and Vs = 35 m/s as car A is the source. And f is the frequency of sound wave at source that is car A.
Similarly, the doppler shift in frequency is the frequency of sound heard by car B which is f' = 1140 Hz. And v is the speed of sound that is v = 343 m/s
1140 = 
f = 1140/1.0649= 1071 Hz.
Thus, the original frequency of horn of Car A is 1071 Hz.
The energy the electron gives up passing between the electrodes is equal to the product between its charge and the potential difference between the electrodes:

where
e is the electron charge

is the potential difference
Plugging numbers into the equation, we find that the electron gives up is