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3 years ago

How much work is done by a crane that lowers 1000N of material a distance of 150

2 answers:

3 0

The crane does NO work.

Work = (force) x (distance)

= (1,000 N) x (150 meters ?) = 150,000 joules

But the work is in the direction of the force, and it isn't the crane
pushing the material down. It's gravity.

Gravity is doing the work.

If the material was being lowered by a cable wrapped around the shaft
of an electric generator, then you could use the work done by gravity to
generate some electrical energy, and then sell the energy.

Or, if the "material" happened to be water, you could let gravity lower it
through a turbine or a water wheel, and use the work done by gravity to
grind wheat.

Yes, the crane may be raising a sweat, working against gravity. The
purpose is only to prevent gravity from doing the work too fast.

Citrus2011 [14]3 years ago

3 0

Since workdone is Force times distance Given force-1000n Distance-150 Wd-1000 times 150 Wd-150000n.

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A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of

Vladimir79 [104]

Answer:

$m = 2.23 \times 10^{-32} kg$

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

orbital speeds = 220 km/s

we know that

acceleration due to centripetal force is $a = \frac{F}{m} = \frac{V^2}{r}$

Gravitational force $F= \frac{Gm m}{d^2}$

we know that

$v = \frac{2\pi R}{T}$

solving for

$R = \frac{vT}{2\pi}$

$F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}$

$F = G\times \frac{\pi m}{(vT)^2}$

$a = \frac{v^2}{\frac{vT}{2\pi}}$

$a = \frac{2\pi v}{T}$

we know that

f =ma

$G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}$

solving for m

$m = \frac{2Tv^3}{\pi G}$

$m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}$

$m = 2.23 \times 10^{-32} kg$

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2 years ago

Objects with unlike charges what each other

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3 years ago

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One 10.0 Ω resistor is wired in series with two 10.0 Ω resistors in parallel. A 45.0 V battery supplies power to the circuit. Wh

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The answer you are looking for 3.0 A

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2 years ago

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I need so much help 50 points!!!!!! complete the graph

motikmotik

First Symbol : Cobalt (Co)

Its Group Number - 9

Its Period Number - 4

Its Family Name - Transition Metal

Second Symbol : Silicon (Si)

Its Group Number - 14

Its Period Number - 2

Its Family Name - Semiconductor

Third Symbol : Astatine (At)

Its Group Number - 17

Its Period Number - 6

Its Family Name - Halogen

Fourth Symbol : Magnesium (Mg)

Its Group Number - 2

Its Period Number - 3

Its Family Name - Alkaline Earth Metal

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Its Group Number - 18

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2 years ago

A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m

Levart [38]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

$a=2.92164g$

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

$\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s$

Angular velocity is 1.3823 rad/s

Linear velocity is given by

$v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s$

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

$a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2$

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

$\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g$

$a=2.92164g$

Centripetal force is given by

$F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N$

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

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3 years ago