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3 years ago

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: A honeybucket man is carrying his load. He has a pole 2 m long with a bucket hanging from each end. The buckets have a mass of

2 kg each. The pole has a mass of 1 kg. The front bucket has 5 kg of "honey" in it, and the rear bucket has 3 kg of "honey." How far from the center of the pole is the honeybucket man's shoulder

1 answer:

nasty-shy [4]3 years ago

8 0

Answer:

Explanation:

Using the principle of moment, assuming the rod is uniform rod of mass 1 kg

the center of mass of the rod will be at 1 m

assuming the system is in equilibrium,

clockwise moment = anticlockwise moment

let the distance of the man shoulder be x from the center of gravity and also is the pivot point

total mass of bucket + mass of honey = 2kg + 3 kg = 5 kg for rear bucket and

2kg + 5 kg = 7 kg for front bucket

( 5kg × ( 1+x)) + ( 1 kg × x) = 7 kg × ( 1 - x)

5 + 5 x + x = 7 - 7x

5 + 6x = 7 - 7x

6x + 7x = 7 - 5

13x = 2

x = 2 / 13 = 0.154 m

the honeybucket man's shoulder is 0.154 m from the center of the pole ( forward ).

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A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

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Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

$h=0+0.5\times 5\times 10^2$

$h=250\ m$

b)

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

<u>height fallen freely by Austin:</u>

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

<u>Velocity just before opening the parachute:</u>

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

<u>Time taken by the helicopter to fall:</u>

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

now,

$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

<u>remaining time,</u>

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

<u>Now the height fallen in the remaining time using parachute:</u>

$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>

$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

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A student is transmitting sound waves through various materials. Through which metal in the table will the sound waves travel th

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Answer:

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3 years ago

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mr_godi [17]

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4 years ago

If Mrs. Reichelt throws a chromebook, because it won't login correctly, with a force of 8N, and the chromebook accelerates at 5m

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Answer:

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Step-by-step Solution:

Since Force = mass × acceleration we have:

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A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.10 m/s. How

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Answer:

769,048.28Joules

Explanation:

A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.10 m/s. How much energy was lost to air friction during this bump

The energy lost due to friction is expressed using the formula;

Energy lost = Potential Energy + Kinetic Energy

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g is the acceleration due to gravity

h is the height

v is the speed

Substitute the given values into the formula;

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