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3 years ago

11

An unfortunate 18 kg monkey falls from a 40 m tall tree. What is the monkeys final velocity just befor he impacts the ground.? a

7063.2 m/s b 784 m/s c 28 m/s d 720 m/s

1 answer:

zalisa [80]3 years ago

5 0

The correct answer is c) 28 m/s.
Let's find the step-by-step solution. The motion of the monkey is an uniformly accelerated motion, with acceleration equal to $g=9.81 m/s^2$ . The initial velocity of the monkey is zero, while the distance covered is S=40 m. Therefore, we can use the following relationship to find vf, the final velocity of the monkey:
$2aS=v_f^2-v_i^2=v_f^2$
from which
$v_f= \sqrt{2aS}= \sqrt{2\cdot 9.81 m/s^2 \cdot 40 m}=28 m/s$

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Which of the following statements is FALSE about collisions? (Consider Newton's Laws an the Conservation of Momentum.)

pantera1 [17]

<u>The following statements are false about collisions: </u>

The velocity change of two respective objects involved in a collision will always be equal.
Total momentum is always conserved between any two objects involved in a collision.

Answer: Option B, and D

<u>Explanation: </u>

In any collisions, equal amount of net force will be acted upon the colliding objects due to the third law of Newton, irrespective of the significance difference in mass of the objects. Similarly, they can also have different acceleration values during collision of two objects if the masses are identical.

But the statements regarding the equal change in velocity of two objects respectively involved in collision always is false, as the conservation of momentum is applicable for isolated system only. So it is true for only isolated system and not in all the systems.

The same reason goes for falsifying the fourth statement which states that total momentum is always conserved between two objects involved in a collision as this statement is only true for isolated system where the conservation of momentum can be applied. Thus the second and fourth statement is false regarding collision.

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3 years ago

Ezra is pulling a sled, filled with snow, by pulling on a rope attached to the sled. The rope makes an angle θ with respect to t

svetoff [14.1K]

Answer:

Explanation:

Given

rope makes an angle of $\theta$

Mass of sled and snow is m

Normal Force $=F_N$

applied Force is F

as Force is pulling in nature therefore normal reaction is given by

$F_N=mg-F\sin \theta$

Also $F\cos \theta =f_r$

$f_r=\mu _k\cdot F_N$

$f_r=\mu _k\cdot (mg-F\sin \theta )$

$F\cos \theta =\mu _kF_N$ -------1

$F\sin \theta =mg-F_N$ ---------2

Squaring 1 & 2 and then adding

$F^2=(\mu _kF_N)^2+(mg-F_N)^2$

$F=\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}$

Substitute value of F in 1

$cos\theta =\frac{\mu _KF_N}{\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}}$

$\theta =cos^{-1}(\frac{\mu _KF_N}{\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}})$

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3 years ago

Which of the following technologies would produce the least energy in light

AysviL [449]

Answer: C. Radio-controlled toy airplane

Explanation:

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3 years ago

In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w

Daniel [21]

Answer:

$v_2=-133.17m/s$ , the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before ( $p_i$ ) and after ( $p_f$ ) the explosion. We will take the east direction as positive.

Before the explosion we have $p_i=m_iv_i=Mv_i$ .

After the explosion we have pieces 1 and 2, so $p_f=m_1v_1+m_2v_2$ .

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

$Mv_i=m_1v_1+m_2v_2$

Which means (since we want $v_2$ and $M=m_1+m_2$ ):

$v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}$

So for our values we have:

$v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s$

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