V = 310 m/s
f = 60 MHz = 60 × 10^6 Hz
v = xf
x = v/f
x = 310/(60 × 10^6) m
x = 5.166667 × 10^(−6) m
I think you forgot to give the choices along with the question. I am answering the question based on my knowledge and research. You would increase mechanical advantage by <span>making the blade longer from the cutting edge. I hope that this is the answer that has actually come to your desired help.</span>
Daniddmelo says it right there, don't know why he got reported.
The potential energy (PE) is mass x height x gravity. So it would be 25 kg x 4 m x 9.8 = 980 joules. The child starts out with 980 joules of potential energy. The kinetic energy (KE) is (1/2) x mass x velocity squared. KE = (1/2) x 25 kg x 5 m/s2 = 312.5 joules. So he ends with 312.5 joules of kinetic energy. The Energy lost to friction = PE - KE. 980- 312.5 = 667.5 joules of energy lost to friction.
Please don't just copy and paste, and thank you Dan cause you practically did it I just... elaborated more? I dunno.
I would say D.) The ball bounces many times suggesting the energy is used up efficiently
