(a) We will use the equation v = u + at
Initial velocity u = 5.00 m/s
Acceleration a = 0.0600 m/s²
time = 8 min = 8 x 60 = 480 s
Final velocity
= u + at
= 5.00 + 0.0600(480)
= 33.8 m/s
The final velocity is 33.8 m/s
Answer:
i think it is car B because the greater the speed, the greater the drag force acting on the vehicle
Explanation:
Explanation:
Using the formula;
2x = vt
x is the distance up from the ocean floor the submarine is
v is the speed of sound in water
t is the time
Given
t = 2.3s
v = 1490m/s
Required
how high (approx) up from the ocean floor is the submarine x
From the formula;
x = vt/2
x = 1490(2.3)/2
x = 745(2.3)
x = 1,713.5m
Hence the submarine is 1713.5m high up from the ocean floor
<span> 0.17 km/m east 0.2 km/m east 5 km/m east 25 km/m east</span>
Logically both masses will collide and well make a reaction. first of all depending on the small mass it will either merge or unite with the big mass or it will bounce away from it . if this happen it will make a reaction that will affect both masses. Hope this helps if it is incorrect please let me know :)
