A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comforta
1 answer:
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>
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Answer:
The displacement of the train in this time period is 2,616.86 m.
Explanation:
A Uniformly Varied Rectilinear Motion is Rectilinear because the mobile moves in a straight line, Uniformly because of there is a magnitude that remains constant (in this case the acceleration) and Varied because the speed varies, the final speed being different from the initial one.
In other words, a motion is uniformly varied rectilinear when the trajectory of the mobile is a straight line and its speed varies the same amount in each unit of time (the speed is constant and the acceleration is variable).
An independent equation of useful time in this type of movement is:
<em>Expression A</em>
where:
- vf = final velocity
- vi = initial velocity
- a = acceleration
- d = distance
The equation of velocity as a function of time in this type of movement is:
vf=vi + a*t
So the velocity can be calculated as:
In this case:
- vf=42.4 m/s
- vi=27.5 m/s
- t=75 s
Replacing in the definition of acceleration:
a=0.199 m/s²
Now, replacing in expression A:
Solving:
d= 2,616.86 m
<u><em>The displacement of the train in this time period is 2,616.86 m.</em></u>
Answer:
15.03 m
Explanation:
Given:
mass of the block, m = 7.25 kg
Angle, Θ = 28.5°
Initial speed of the block, v₀ = 15 m/s
let the distance traveled by the block be 's'
Now, applying the work energy theorem,
we have
on substituting the values in the above equation, we get
or
or
s = 15.03 m
Hence, the block will travel 15.03 m up the ramp