Answer:T=1316.21 N
Explanation:
The tension has two components: Vertical and Horizontal. The
horizontal component is ma, the vertical component is mg. Using
Pythagoras theorem, we can find the tension as:
T=((ma)^2 (mg)^2)^(1/2)
So
T=((129*2.84)^2 (129*9.8)^2)^(1/2)
T=1316.21 N
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:
(C) Average Velocity = 25 m/s due North
Answer:
The initial velocity was U=22.14m/s
Explanation:
Step one :
Applying the third equation of motion
v² = u²+ 2as
Where v= Final velocity
U =initial velocity
a= acceleration due to gravity
S= distance or displacement
Step two :
V= 0
a= 9.81m/s²
S=25m
U=?
Step three :
Substituting into the equation we have
0²=U²+2*9.81*25
0=U²+490.5
U²=-490.5
U=√490.5
U=22.14m/s