The velocity of the submarine immediately after firing the missile is 0.0104 m/s
Explanation:
Mass of the submarine M=50 tonne=
Mass of the missile m=40 kg
velocity of the missile v= 13m/s
we have to calculate the velocity of the submarine after firing
This is the recoil velocity and its expression is derived from the law of conservation of momentum
recoil velocity of the submarine
Answer:
acceleration, a = 9.8 m/s²
Explanation:
'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.
u = 0 m/s
After 2 seconds, velocity of the ball is 19.6 m/s.
t = 2s, v = 19.6 m/s
Using
v = u + at
19.6 = 0 + 2a
a = 9.8 m/s²
yes it does it because the material cause it to refract and the stronger it is the more it will refract
At the bottom of the tank :
P = ρgH
P = (1000 kg/m³)(10 m/s²)(1 m)
P = 10000 N/m²
F = P • A
F = (10000 N/m²)(1 m²)
F = 10000 N
At the side of the tank :
Pav = ½ρgH
Pav = ½(1000 kg/m³)(10 m/s²)(1 m)
Pav = 5000 N/m²
F = P • A
F = (5000 N/m²)(1 m²)
F = 5000 N
The answer should be B. a stable isotope to a decaying isotope.
