X component of force Fx=Fcos45=50×cos×45
Y component of force Fy=Fsin45=50×sin×45
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:
Explanation:
Total heat = Work done = Force × distance
distance = 0.075 × 12 = 0.9 m
W = 45 × 0.9 = 40.5 joules
Specific heat of the human hand = 3.5 kj/kg = 3.5 j/g
Q = MCΔT
ΔT = (Q) ÷ (MC)
ΔT = 40.5 ÷ (3.5 × 1) = 11.57°C
Explanation:
its the minimum amount of energy required to remove the most loosely bound electron
Answer:
okay here is a thing I learned when I was younger in my middle school:
Explanation:
my teacher would tell me that metals are considered a weak metals are on the left side and the good metals are located on the right side because the only way I remembered was the right means it is really strong and the left is weak and not that supportive. but I think that's how I still think it is or other people may have their own opinions. but hope this helped out with your question!
