Uploaded by Khoa Lê Trần Đăng

REPORT LAB 1 (1)

advertisement
REPORT LAB 1:
DETERMINING DENSITY OF RIGID OBJECTS
Class: …….…. / Group: ……Team:……….
Lecturer’s comment
Full name:
1) ………………………………………......
2) ………………………………………......
3) ………………………………………......
4) ………………………………………......
5) ………………………………………......
I. Aims/Purposes
.................................................................................................................
.................................................................................................................
II. Apparatus, Methods, and Procedure
.................................................................................................................
.................................................................................................................
III. Equations
.................................................................................................................
.................................................................................................................
IV. Experimental data
Table 1:
Accuracy of Vernier caliper: ....................... (mm)
D
(10-3 m)
ΔD
(10-3 m)
d
-3
(10 m)
Δd
(10-3 m)
h
-3
(10 m)
Δh
(10-3 m)
1
2
3
Avg.
Table 2:
Accuracy of Vernier caliper: ....................... (mm)
a
-3
(10 m)
Δa
(10-3 m)
b
-3
(10 m)
Δb
(10-3 m)
c
-3
(10 m)
Δc
(10-3 m)
1
2
3
Avg.
Table 3:
Accuracy of Vernier caliper: ....................... (mm)
1
2
3
Average
=............... (10-3 m)
(10-3 m)
=.............. (10-3 m)
-3
(10 m)
Table 4: The mass of the ring, the cube, and the ball
Accuracy of technical balance: …………………..(g)
m1
m1
m2
m2
m3
m3
(103 kg ) (103 kg ) (103 kg ) (103 kg ) (103 kg ) (103 kg )
1
2
3
Avg
.
V. Calculations
V.1 Calculating the absolute errors of the measurement of dimensions
∆𝑥 = ∆𝑥𝑖𝑛𝑠𝑡𝑟𝑢𝑚𝑒𝑛𝑡 + ∆𝑥 (𝑚)
V.2 Calculating the total absolute errors of the measurement of mass m
∆𝑚 = ∆𝑚𝑠𝑦𝑠 + ∆𝑚 (𝑘𝑔)
V.3 Calculating the density of rigid objects
(𝜋 = 3.14 and ∆𝜋 = 0.005)
𝑉1 =
𝜋 2
1
(𝐷 − 𝑑 2 )ℎ (𝑚3 ), 𝑉2 = 𝑎 × 𝑏 × 𝑐 (𝑚3 ), 𝑉3 = 𝜋𝑑 3 (𝑚3 )
4
6
and
∆𝑉1
𝑉
=
1
∆𝜋
+
2(𝐷. ∆𝐷 + 𝑑. ∆𝑑)
2
+
∆ℎ
ℎ
𝐷2 − 𝑑
∆𝑉2 ∆𝑎 ∆𝑏 ∆𝑐
=
+
+
=𝛿
𝑉
𝑎
𝑏
𝑐
𝜋
=𝛿
2
∆𝑉3
𝑉
=
3
∆𝜋
𝜋
+
3∆𝐷
𝐷
=> ∆𝑉 = 𝛿. 𝑉 = ⋯ … … (𝑚3 )
Thus,
𝜌=
𝑚
𝑉
= ⋯……(
𝑘𝑔
𝑚3
)
∆𝜌 ∆𝑚 ∆𝑉
=
+
=𝛾
𝜌
𝑚
𝑉
=> ∆𝜌 = 𝜌. 𝛾 = ⋯ … … (
𝑘𝑔
𝑚3
)
VI. Conclusion
𝜌 = (𝜌 ± ∆𝜌) = ⋯ … … … … … (
𝑘𝑔
)
𝑚3
Download