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ankoles [38]
4 years ago
11

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

Physics
2 answers:
ludmilkaskok [199]4 years ago
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]4 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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Answer:

Explanation:

The forces compare together as a result of the fact that the force exerted by that of the ball and the force exerted by that of the wall both have the same magnitude.

8 0
3 years ago
Friction occurs when the and of two surfaces stick to each other
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yes, kinda like when babies are made, there is friction

Explanation:

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A charge Q is distributed uniformly around the perimeter of a ring of radius R. Determine the electric potential difference betw
il63 [147K]

Answer:

the electric potential difference between the point at the center of the ring and a point on its axis ΔV is ( 0.8356 )\frac{kQ}{R}

Explanation:

Given the data in the question;

electric potential at the center of the ring V₀ = kQ / R

electric potential on the axis point Vr = kQ / √( R² + x² )

at a distance 6R from the center,

point at x = 6R

so distance circumference r = √( R² + (6R)² )

so

electric potential on the axis point Vr = kQ / √( R² + (6R)² )

Vr = kQ / R√37

Now

ΔV = V₀ - Vr

we substitute

ΔV = ( kQ / R) - ( kQ / R√37 )

ΔV =  kQ/R( 1 - 1/√37 )

ΔV =  kQ/R( 1 - 0.164398987 )

ΔV =  kQ/R( 0.8356 )

ΔV = ( 0.8356 )\frac{kQ}{R}       { where k = \frac{1}{4\pi e_0} }

Therefore, the electric potential difference between the point at the center of the ring and a point on its axis ΔV is ( 0.8356 )\frac{kQ}{R}

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Answer:

The Answer is A

Explanation:

4 0
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Suppose a car's standard tires are replaced with tires 1.30 times larger in diameter.(i) Will the car's speedometer reading be (
Nezavi [6.7K]

The answer is 1.30 times too low.

<h3>What is a speedometer reading?</h3>
  • According to the law, speedometers can never understate a vehicle's speed while never exceeding 110% of the real speed + 6.25 mph. Therefore, if you're traveling at 40 mph, your speedometer may display up to 50.25 mph, but it will never show a speed below 40 mph.
  • While the speedometer displays how fast your car is moving, the odometer displays how far it has traveled. According to Tetzlaff, the speedometer's accuracy in most cars, including Volkswagens, is typically within a few percentage points of actual speed.
  • A car's speedometer is a tool that displays the vehicle's speed. An odometer is yet another accessory connected to a speedometer. The car's odometer displays the distance driven. The speedometer, as we all know, displays the pace of every single moment.
  • Federal law prohibits speedometers from having an inaccuracy of more than 5% (usually stated as plus/minus 2.5% relative to the real speed), according to Dan Edmunds, an automotive engineer and director of vehicle testing at Edmunds.com.

The answer is 1.30 times too low.

To learn more about speedometers, refer to:

brainly.com/question/3397467

#SPJ4

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2 years ago
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