Response:
2.5kN.m
Details:
Torque relates directly to the pitch diameter
= Ta/Tb= Da/Db
For 120/Tb= 0.25/0.5
This gives Tb= 2.469kN.m, roughly 2.5kN.m
Response:
C. vx
F. ax
G. ay
Clarification:
The projectile follows a curved trajectory toward the ground, causing changes in x and y positions.
Since there is no external force acting in the x-direction, the acceleration in x remains at zero. Consequently, ax and vx remain unchanged.
The projectile is subject to the force of gravity, directed downwards, leading to an increase in its velocity due to the rise in its y-component.
Meanwhile, the y-component of acceleration remains constant due to gravitational acceleration.
Answer
Ceres, Pluto, and Eris are categorized as DWARF PLANETS.
A) Remaining planetesimals formed within the frost line are referred to as ASTEROIDS.
B) METEORITES are fragments of asteroids that have landed on Earth.
C) COMETS are celestial objects that are often visible with their long tails.
D) COMETS are also planetesimals that were left over and originated in the region of the solar system dominated by the jovian planets.
E) Meteor showers are linked to debris from COMETS.
Answer:
F=126339.5N
Explanation:
To compute the force required to escape, a free-body diagram for the hatch must be drawn. We will equate the downward and upward forces, thus applying the following equation:
Fw=W+Fi+F
where
Fw= force or weight exerted by the water column above the submarine.
To calculate Fw, we can use:
Fw=h. γ. A
h=height
γ=
specific weight of seawater = 10074N / m ^ 3
A=Area
Fw=28x10074x0.7=197467N
w represents the hatch weight = 200N
Fi denotes the internal pressure force in the submarine, which is 1 atm = 101325Pa. We can calculate this force using:
Fi=PA=101325x0.7=70927.5N
Finally, the force needed to open the hatch is determined by the original equation:
Fw=W+Fi+F
F=Fw-W+Fi
F=197467N-200N-70927.5N
F=126339.5N
