Answer:
Maximum force, F = 1809.55 N
Explanation:
Given that,
Diameter of the anterior cruciate ligament, d = 4.8 mm
Radius, r = 2.4 mm
The tensile strength of the anterior cruciate ligament,
We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,
So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N
Answer:
B) 2I
Explanation:
The equation that relates voltage, current and resistance is V=RI.
The equation for the resistance of a material in terms of its resistivity, length and cross-sectional area is
In this case, the length is divided by 2 while keeping its resistivity (since it's the same material) and area, which means the resistance gets divided by 2. Then, looking at the equation I=V/R and keeping V constant, one deduces that since the resistance now is half than before then current now must be twice as before.
This is all intuitive in fact, cuting a homogeneous resistor in half and leaving the rest of the variables constant makes twice as easy for the electrons to cross the conductor, thus twice the current (one has to know that all the variables involved behave linearly, as the equations show).
Answer:
442.36038 m or 1451.31362 ft
Explanation:
= Initial pressure = 30.15 inHg
= Final pressure = 28.607 inHg
= Density of air = 0.075 lb/ft³
Density of mercury = 13560 kg/m³
g = Acceleration due to gravity = 9.81 m/s²
Difference in pressure is given by
The height of the building is 442.36038 m or 1451.31362 ft