Assignment 3
Harprit Prashant Wadekar
(1)
The diameter of piston and cylinder = 40 mm(D)
Pressure threshold ≤ 3500
Force (F) = Pressure (P) × Area (A)
=
To find Area (A) =
= 1256.64 mm2
F = P × A = 3500 × 1256.64 = 4398240 N
The maximum force that can withstand under 3500kPa is 4398240 N
(2)
2 ln(𝑒
)−𝑒
(
)
=2(3𝑥 + 1) ln(𝑒) − 𝑒
=(6𝑥 + 2)(1) − 𝑒
(
(
)
(Power rule)
)
= 6𝑥 + 2 − (2𝑥 − 3)
(𝑒
( )
= 𝑚)
= 6𝑥 + 2 − 2𝑥 + 3
= 𝟒𝒙 + 𝟓
Solve for y in terms of x
2 ln 𝑥 + ln 𝑦 = 1 + ln 5
ln 𝑦 = 1 + ln 5 − 2 ln 𝑥
𝑒
= 𝑒(
)
𝑦 = 𝑒(
)
𝑦 = 𝑒 × 𝑒(
)
× 𝑒(
𝑦 = 𝑒 × 𝑒(
)
× 𝑒(
)
)
𝑦 = 𝑒×5×𝑥
𝒚=
𝟓𝒆
𝒙𝟐
(3)
Given value of resistance(R)=12Ω
The power dissipa on (p) which varies from 2.50w to 8w
Power dissipa on (p) = 𝑅𝑖
𝑖=
𝑝
𝑅
For p=2.50w and R=12Ω
𝑖=
𝑖=
2.50
12
𝑖 = 0.457 𝐴
For p=8w and R=12Ω
𝑖=
𝑖=
8
12
𝑖 = 0.82 𝐴
The current varies between 0.457 A and 0.82 A