The force applied to small piston = 2.2 x 10³ N
<h3>Further explanation</h3>
Given
a radius of 5 cm and 15 cm
weight 20000 N
Required
Force applied
Solution
Pascal Law :
F₁/A₁=F₂/A₂
A₁ = π.5²
A₂ = π.15²
F₁/ π.5² cm² = 20000/π.15² cm²
F₁ = 2222.22 N⇒2.2 x 10³ N
Answer: Actually three of them are. The ovaries, the uterus and fallopian tubes.
Answer:
The friction coefficient's minimum value will be "0.173".
Explanation:
The given query seems to be incomplete. Below is the attached file of the complete question.
According to the question,
(a)
The net friction force's magnitude will be:
⇒ 


(b)
For m₃,
⇒ 
Or,
⇒ 


Does that help I hope IT does you probably just have to write the 1st sentence.