Question

A neutron moving with a constant speed passes two points 3.6 m apart in 1.8 × 10^-4s.The kinetic energy of neutron is​

Answer 1

Given that,

Distance = 3.6 m

Time $t=1.8\times10^{-4}\ sec$

We need to calculate the velocity

Using formula of velocity

$v=\dfrac{d}{t}$

Put the value into the formula

$v=\dfrac{3.6}{1.8\times10^{-4}}$

$v=20000\ m/s$

We need to calculate the kinetic energy

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(20000)^2$

$K.E=3.34\times10^{-19}\ J$

Hence, The kinetic energy of neutron is $3.34\times10^{-19}\ J$

Answer 2

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