Answer:
The K.E of the bowling ball right before it hits the ground, K.E = 2450 J
Explanation:
Given data,
The mass of the bowling ball, m = 10 kg
The height of the building, h = 25 m
The total mechanical energy of the body is given by,
E = P.E + K.E
At height 'h' the P.E is maximum and the K.E is zero,
According to the law of conservation of energy, the K.E at the ground before hitting the ground is equal to the P.E at 'h'
Therefore, P.E at 'h'
P.E = mgh
= 10 x 9.8 x 25
= 2450 J
Hence, the K.E of the bowling ball right before it hits the ground, K.E = 2450 J
Answer:
your right answer is true
hope it helps you
The frequency of bird chirping hear by hiran will be 1.77 kHz.
<u>Explanation:</u>
As per Doppler effect, the observer will feel a decrease in the frequency of the receiving signal if the source is moving away from the observer. So the shifted frequency is obtained using the below equation:

Here , c is the speed of sound, Vs is the velocity of source with which it is moving away. f is the original frequency of source and f' is the frequency shift heard by the observer.
As here, f = 1800 Hz, Vs= 6 m/s and c = 343 m/s, then

So, the frequency of bird chirping hear by hiran will be 1.77 kHz.
Increases exponentially is your correct answer
Answer:
it is because it do not produce smoke like fuel from wood.