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2 years ago

13

A person holds a portable fire extinguisher that ejects 1.0kg of water per second horizontally at a speed of 6.0 m/s. What newto

ns must the person exert on the extinguisher in order to prevent it from accelerating?
(A) 0N
(B) 6N
(C) I0N
(D) 18N
(E) 36N

1 answer:

Leviafan [203]2 years ago

7 0

Answer:

6 N

Explanation:

$\dfrac{m}{t}$ = Mass flow rate = 1 kg/s

v = Final velocity = 6 m/s

u = Initial velocity = 0 m/s

Force is obtained when we divide change in momentum by time

$F=\dfrac{m(v-u)}{t}\\\Rightarrow F=\dfrac{m}{t}(v-u)\\\Rightarrow F=1(6-0)\\\Rightarrow F=6\ N$

The force the person exert on the extinguisher in order to prevent it from accelerating is 6 N

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Answer:

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Let $t_2-t_1 = \Delta{t}$ , then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

$v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}$

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Solving the system of equations:

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$v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s$

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2 years ago