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3 years ago

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The high-speed winds around a tornado can drive projectiles into trees, building walls, and even metal traffic signs. In a labor

atory simulation, a standard wood toothpick was shot by pneumatic gun into an oak branch. The toothpick mass was 0.12 g, its speed before entering the branch was 221 m/s, and its penetration depth was 16 mm. If its speed was decreased at a uniform rate, what was the magnitude of the force of the branch on the toothpick?

1 answer:

Travka [436]3 years ago

3 0

Answer:

F = 183.153 N

Explanation:

given,

mass of the toothpick = 0.12 g = 0.00012 kg

initial velocity = 227 m/s

final velocity = 0 m/s

penetration depth = 16 mm = 0.016 m

using the equation of motion

v² - u² = 2 a s

0 - u² = 2 a s

- 221² = 2 × a × 0.016

a = 1526281.25 m/s²

Force is equal to

F = m a

= 0.00012 × 1526281.25

F = 183.153 N

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P=9.58 W

Explanation:

According to Newton's second law, and assuming friction force as zero:

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$a=\frac{\Delta v}{t}\\a=\frac{0.685m/s}{21.5*10^{-3}s}\\\\a=31.9m/s^2$

So the force exerted by the motor is:

$F_m=0.875kg*31.9m/s^2\\F_m=27.9N$

The work done by the motor is given by:

$W_m=F_m*d\\\\d=\frac{1}{2}*a*t^2\\d=\frac{1}{2}*31.9m/s^2*(21.5*10^{-3}s)^2\\\\d=7.37*10^{-3}m$

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And finally, the power is given by:

$P=\frac{W_m}{t}\\P=\frac{0.206J}{21.5*10^{-3}s}\\\\P=9.58W$

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Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

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3 years ago