Answer:
<em>The depth will be equal to</em> <em>6141.96 m</em>
<em></em>
Explanation:
pressure on the submarine 
= 62 MPa = 62 x 10^6 Pa
we also know that = ρgh
where
ρ is the density of sea water = 1029 kg/m^3
g is acceleration due to gravity = 9.81 m/s^2
h is the depth below the water that this pressure acts
substituting values, we have
= 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine 
= 101 kPa = 101000 Pa
this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.
Equating the pressure , we have
62 x 10^6 = 10094.49h
depth h = <em>6141.96 m</em>
Answer:
50m
Explanation:
Given parameters:
Initial velocity = 20m/s
Acceleration = 4m/s²
Time = 10s
Unknown:
Distance traveled by the rocket = ?
Solution:
To solve this problem use the expression below;
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
final velocity = 0
Insert the parameters and solve;
0² = 20² + 2 x 4 x s
-400 = 8s
s = 50m
Disregard the negative sign because distance cannot be negative.
Hello! Assuming that the only force acting on the mass is 30N...
Fnet = 30N
Fnet = ma (mass x acceleration)
ma = 30N
a = 30N / m
a = 30N / 7kg
a = 4.2857 m/s^2
a = 4 m/s^2
I hope this helps!
The answer would be, "1/560 seconds".
