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borishaifa [10]
3 years ago
12

A train starts from rest and accelerates uniformly, until it has traveled 3.7 km and acquired a velocity of 26 m/s. What is the

acceleration of the train during this time?
A. 0.12 m/s^2
B. 0.082 m/s^2
C. 0.11 m/s^2
D. 0.091 m/s^2
E. 0.10 m/s^2
Physics
1 answer:
bogdanovich [222]3 years ago
7 0
We have:
s=3,7km (3700m)
u=0
v=26m/s
a=
t=
We look up for a formula to solve for a:
v²=u²+2as u²=0 so
v²=2as
(26m/s)²=2a3700m
676/7400=a
a=0,09m/s²

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You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

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The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

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