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miabarrett8117
21.03.2020 •
Physics
A 0.43 kg hammer is moving horizontally at 5.2 m/s when it strikes a nail and comes to rest after driving the nail 0.014 m into a board. What is the duration (in seconds) of the impact?
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Ответ:
5.385×10⁻³
Explanation:
First we find the acceleration
Using the equation of motion,
v² = u²+2as Equation 1
Where v = Final velocity, u = initial velocity, a = acceleration, s = distance.
make a the subject of the equation,
a = (v²-u²)/2s Equation 2
Given: v = 0 m/s (comes to rest), u = 5.2 m/s, s = 0.014 m
Substitute into equation 2
a = (0²-5.2²)/(2×0.014)
a = -27.04/0.028
a = -965.71 m/s²
Finally Using
a = (v-u)/t
where t = Duration of impact
make t the subject of the equation
t = (v-u)/a Equation 3
Given: v = 0 m/s, u = 5.2 m/s, a = -965.71 m/s²
Substitute into equation 3
t = (0-5.2)/-965.71
t = -5.2/-965.71
t = 5.385×10⁻³ s.
Hence the duration of impact = 5.385×10⁻³
Ответ: