How do I solve for part (c)? Please help! Will rate lifesaver if correct! A grou
ID: 2115300 • Letter: H
Question
How do I solve for part (c)? Please help! Will rate lifesaver if correct!
A group of students performed the same "Newton's Second Law" experiment that you did in class. For this lab, assume g = 9.81 m/s2. They obtained the following results:
m1(kg)
t1(s)
v1(m/s)
t2(s)
v2(m/s)
0.050
1.2000
0.2500
1.7880
0.4102
0.100
1.2300
0.3240
1.6296
0.6570
0.150
1.1500
0.3820
1.4735
0.8845
0.200
1.1100
0.4240
1.3914
0.9674
where m1 is the value of the hanging mass (including the mass of the hanger), v1 is the average velocity and t1 is the time at which v1 is the instantaneous velocity for the first photogate, and v2 is the average velocity and t2 is the time at which v2 is the instantaneous velocity for the second photogate.
(a)
slope = 0.8493 kg
y - intercept = 0.2504 N
(b)
Total mass of the system M = 0.8493 kg
(c) Using the information you obtained in parts (a) and (b), predict what the value of the acceleration would be if the value of the hanging mass were increased to m1 = 0.35 kg.
m1(kg)
t1(s)
v1(m/s)
t2(s)
v2(m/s)
0.050
1.2000
0.2500
1.7880
0.4102
0.100
1.2300
0.3240
1.6296
0.6570
0.150
1.1500
0.3820
1.4735
0.8845
0.200
1.1100
0.4240
1.3914
0.9674
Explanation / Answer
using y=mx+c
equation is F=0.8493a+0.2504
substutute F=m1g=0.35*9.8=3.43 you get
3.43= 0.8493*a+0.2504 or
a = 3.743m/s2
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