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avanturin [10]
2 years ago
13

Calculate the work done to push a 200-N object 5-meters

Physics
1 answer:
AVprozaik [17]2 years ago
6 0
1000 joules
jskskmxk dnd
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Pertaining to simple machines and levers what changes when the fulcrum position is modified?
Inessa [10]

Explanation :

There are many types of simple machines i.e. lever, wedge, pulley, wheel and axle etc.

Fulcrum is the pivot point about which the lever turns. When the position of the fulcrum is shifted, the position of effort changes.

There are three classes of the lever as class 1, class 2 and class 3.

Class 1 includes seesaws and scissors.

class 2 includes wheelbarrow

class 3 includes tongs and tweezers.

7 0
3 years ago
You are an engineer helping to design a roller coaster that carries passengers down a steep track and around a vertical loop. Th
vova2212 [387]

Answer:

h >5/2r

Explanation:

This problem involves the application of the concepts of force and the work-energy theorem.

The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.

Vertically two forces are acting on the rider, the weight and normal force of the cart on the rider. The normal force and weight are acting downwards at the top. For the rider not to fall out of the cart at the top, the normal force on the rider must be zero. This brings in a design requirement for the roller coaster to move at a minimum speed such that the cart exerts no force on the rider. This speed occurs when the normal force acting on the rider is zero (only the weight of the rider is acting on the rider)

So from newton's second law of motion,

W – N = mv²/r

N = normal force = 0

W = mg

mg = ma = mv²/r

mg = mv²/r

v²= rg

v = √(rg)

The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop

So

ΔPE = ΔKE = 1/2mv²

The height at the roller coaster starts is usually higher than the top of the loop by design. So

ΔPE =mgh - mg×2r = mg(h – 2r)

2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.

In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.

So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.

ΔPE > ΔKE

mg(h–2r) > 1/2mv²

g(h–2r) > 1/2(√(rg))²

g(h–2r) > 1/2×rg

h–2r > 1/2×r

h > 2r + 1/2r

h > 5/2r

5 0
3 years ago
Read 2 more answers
Describe the motion of an unanchored rowboat when a water wave passes
USPshnik [31]
It mimics the movement of the waves 
6 0
3 years ago
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Does a thicker core make an electromagnet stronger?
goldenfox [79]

Answer:

Yes

Explanation:

The core of an electromagnet serves to stabilize the magnetic field created by the wire. The thicker the core, the more metal there is to amplify the current. Therefore, a thicker core does make an electromagnet stronger. Hope this helps!

5 0
3 years ago
A box is being pulled to the right. The free body diagram is shown.What is the magnitude of the kinetic frictional force?
kirill115 [55]

Answer:

A - 0 N

plz mark 5 stars, thanks, and brainliest

4 0
3 years ago
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