When the computer is on the circuit has positive and negative charges it will move fluenlty to get the power when it is in on position.
Answer:
Magnitude of the third vector: 57.85 cm
The direction of the third vector: 44.76 N of W
Explanation:
You said "<span>A rocket's acceleration is 6.0 m/s2.".
That just means that its speed increases by 6 m/s every second.
Whenever you look at it, its speed is 6 m/s faster than it was
one second earlier.
If it starts out with zero speed, then its speed is 6 m/s after 1 second,
12 m/s after 2 seconds, 18 m/s after 3 seconds . . . etc.
How long does it take to reach 42 m/s ?
Well, how many times does it have to go 6 m/s FASTER
in order to build up to 42 m/s ?
That's just (42/6) = 7 times.
Writing it correctly, with the units and everything, it looks like this:
(42 m/s) / (6 m/s</span>²)
= (42/6) (m/s) / (m/s²)
= (42/6) (m/s · s²/m)
= 7 seconds
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>
40 seconds I’m pretty sure sorry if I’m wrong
