Physics Lab about Centripetol Force (basic)

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Science

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Follow the instructions on the Lab and use the data I upload and finish the lab.

Answer on the Lab instruction is okay.

Please use also another paper to do the calculations. [ I want the answer on the lab instruction(including calculation) and the calculation on another paper]

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Physics lab Student’s name Institution 1.6 spring force = - mg Where m=mass G=acceleration force F= - mg =mv^2r V={(277+268+279)/3}/30 V=9.2 =153*(55/100)^2*9.2 F=4,257,990/10000N F=425.799N 2.13 average= (277+279+268)/3 =824/3 =274.67 2.14 period (T)=1/f 274.67/30 =9.2 T=1/9.2 T=0.12s 2.16 V=(2*22/7)0.12 V=0.75 F= (153*0.55^2)/0.75 F=303.71 3.1 centripetal force 3.2 add the amount of mass 3.3 no because when rotating more force is required. 3.4 (425.799-303.73)/425*100 =28.72. Data Lab1 Mass = 153g Radius = 95.0mm Number of truns in 30 s: First Try: 268 Second Try: 277 Thrid Try: 279 Lab 2 Hanging mass: 2480g For lab 1&2 Same object, same spring.
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Explanation & Answer

Attached.

Running head: PHYSICS LAB

Physics lab
Student’s name
Institution Affiliation
Date

PHYSICS LAB

1.6 spring force = - mg
Where m=mass
G=acceleration force
F= - mg
=mv^2r
V={(277+268+279)/3}/30
V=9.2
=153*(55/100)^2*9.2
F=4,257,990/10000N
F=425.799N

2.13 average= (277+279+268)/3
=824/3
=274.67

2.14

period (T)=1/f

274.67/30
=9.2
T=1/9.2
T=0.12s

2.16

V=(2*22/7)0.12

PHYSICS LAB
V=0.75
F= (153*0.55^2)/0.75
F=303.71

3.1

centripetal force

3.2 add the amount of mass
...


Anonymous
Awesome! Perfect study aid.

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