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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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Answer:

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Explanation:

It is given that,

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\lambda\propto \dfrac{1}{T}

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\lambda=\dfrac{b}{T}

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\lambda=\dfrac{2.93\times 10^{-3}}{T}

T=\dfrac{2.93\times 10^{-3}}{\lambda}

T=\dfrac{2.93\times 10^{-3}}{1.9\times 10^{-6}\ m}

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Explanation:

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Length of wire, L = 2 cm = 0.02 m

Radius of the wire, r = 2 mm = 2 X 10⁻³m

Cross section, 3 ms

charge drifts, q = ?

We know,

the charge drifts through the copper wire is given by

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So, i = \frac{\pi(r)^{2}V  }{pL}

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Substituting the values,

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