<h3>Answer;</h3>
<em>B.)neither longitudinal nor transverse</em>
<h3><u>Explanation;</u></h3>
- <em><u>Longitudinal waves</u></em> are waves in which the vibration of particles is parallel to the direction of the wave motion.
- <em><u>Transverse waves</u></em> on the other hand are those waves in which the vibration of particles is perpendicular to the direction of the wave motion.
- In <em><u>surface waves particles in the medium of transmission move in a circular motion.</u></em> Therefore, they are neither transverse waves nor longitudinal waves.
Answer:
technically yes
Explanation:
with a gun depending on how fast it shoots so when you fire at something you shoot in front of it a little bit so you hit it but a plane that fast you shoot like 100 feet infront of it...
If net external force acting on the system is zero, momentum is conserved. That means, initial and final momentum are same → total momentum of the system is zero.
<span>Px = 0
Py = 2mV
second, Px = mVcosφ
Py = –mVsinφ
add the components
Rx = mVcosφ
Ry = 2mV – mVsinφ
Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²)
and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
simplifying
Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²)
Vf = (V/3)âš((cosφ)² + (2 – sinφ)²)
Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ))
Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ))
using the identity sin²(Ď)+cos²(Ď) = 1
Vf = (V/3)âš1 + 4 – 2sinφ)
Vf = (V/3)âš(5 – 2sinφ)</span>
