Answer:
Vy = 26 m/s sin 30 = 13 m/s vertical speed
t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec time to reach Vy = 0
H = Vy t + 1/2 g t^2
H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m
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
The interaction between two like-charged objects is repulsive. ... Positively charged objects and neutral objects attract each other; and negatively charged objects and neutral objects attract each other.
Answer:
A, The same amount of gravity
Explanation:
300 000 0 squared = 2 x 9.8 distance
KINEMATICS
Uniform or constant motion in a straight line (rectilinear). Speed or velocity constant and/or acceleration constant. If motion is up and down and/or has an up and down component then acceleration omn earth will be g. g is about 10m/s/s.
speed = distance/time
velocity = displacement/time
s=distance ... u=initial speed ... v = final speed ... a = acceleration ... t = time
v=u+at
v^2=u^2+2as
s=ut+1/2at^2
