Question

A 2.5 g marshmallow is placed in one end of a 40 cm pipe, as shown in the figure above. A person blows into the left end of the pipe to eject the marshmallow from the right end. The average net force exerted on the marshmallow while it is in the pipe is 0.7 N. The speed of the marshmallow as it leaves the pipe is most nearly

Answer 1

Answer:

15 m/s

Explanation:

Given:

F = 0.7N   m = 0.25g = 0.0025kg   x = 40cm = 0.4m $v_{0}$ = 0

Since F = ma, a = F/m = 0.7N/0.0025kg = 280m/$s^{2}$

Using the Big Five, $v ^{2} = v_0^{2} + 2ax$, $v = \sqrt{0 + 2*280m/s^2*0.4m} = \sqrt{216} m/s= 15m/s$

Answer 2

We have that for the Question it can be said that  the speed of the marshmallow as it leaves the pipe is most nearly

v=15m/s

From the question we are told

A 2.5 g marshmallow is placed in one end of a 40 cm pipe, as shown in the figure above. A person blows into the left end of the pipe to eject the marshmallow from the right end. The average net force exerted on the marshmallow while it is in the pipe is 0.7 N. The speed of the marshmallow as it leaves the pipe is most nearly

Generally the equation for the acceleration  is mathematically given as

$a=\frac{Force}{mass}\\\\a=\frac{F}{m}\\\\Therefore\\\\a=\frac{0.75}{0.0025}\\\\a=280m/s^2$

Using Newtons

V^2=u^2+2as

V^2=2(280)*0.4

v=15m/s

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