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Molodets [167]
3 years ago
14

A heavy rope, 80 ft long and weighing 32 lbs, hangs over the edge of a building 100 ft high. how much work w is done in pulling

the rope up 20 ft?
Physics
1 answer:
allochka39001 [22]3 years ago
6 0
The first thing you should know for this case is that work is defined as the product of force by the distance traveled in the direction of force.
 We have then:
 W = Fd
 The distance varies, so we must integrate:
 from 0 to 20:
 W = ∫F (x) dx
 W = ∫32xdx
 W = 32∫xdx
 W = 32 (x ^ 2/2) = (16) (20 ^ 2) = 6400 ft * lbs
 answer:
 6400 ft * lbs is work done pulling the rope up 20 ft
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There are screws all around you. Name five examples of screws that you see in everyday life? Think broadly!
Ira Lisetskai [31]

Answer:

Some examples of the uses of a screw are in a jar lid, a drill, a bolt, a light bulb, faucets, bottle caps and ball point pens.

Explanation:

3 0
2 years ago
Which of the following are true? Select all that apply. The net electric field at any location inside a block of copper is zero
Agata [3.3K]

Answer:

1) The net electric field at any location inside a block of copper is zero if the copper block is in equilibrium.

2) In equilibrium, there is no net flow of mobile charged particles inside a conductor.

3) If the net electric field at a particular location inside a piece of metal is not zero, the metal is not in equilibrium.

Explanation:

1) and 3) A block of copper is a conductor. The charged particles on a conductor in equilibrium are at rest, so the intensity of the electric field at all interior points of the conductor is zero, otherwise, the charges would move resulting in an electric current.

2) The charged particles on a conductor in equilibrium are at rest.

6 0
3 years ago
A plane designed for vertical takeoff has a mass of 8.0 × 10³ kg. Find the net work done by all forces on the plane as it accele
artcher [175]

Answer:

<em>the net work done after starting from rest is =  2.4 × 10⁵ J</em>

Explanation:

Work: Work can be defined as the product of force and distance. The fundamental unit of work is Joules (J),  The unit of Energy is Joules (J), as such Energy and work are interchangeable during calculation, This is illustrated below

E = W = 1/2mv².......................... Equation 1

Where m = mass of the plane, v = velocity of the plane, E = Energy, W = work done.

v² = u² + 2as ................................. Equation 2.

Where v = final velocity of the plane, u = initial velocity of the plane, a = acceleration of the plane, distance of the plane.

<em>Given: a = 1.0 m/s², s = 30 m, u = 0 m/s (at rest)</em>

<em>Substituting these values into equation 2</em>

<em>v² = 0² +2×1×30</em>

<em>v² = 60</em>

<em>v = √60</em>

<em>v = 7.75 m/s</em>

Also given: m = 8.0 × 10³ kg, and v = 7.75 m/s

<em>Substituting these values into equation 1,</em>

<em>W = 1/2(8.0×10³)(7.75)²</em>

<em>W = (4.0×10³)(60)</em>

W = 240 × 10³ J

<em>W = 2.4 × 10⁵ J</em>

<em>Therefore the net work done after starting from rest is =  2.4 × 10⁵ J</em>

4 0
3 years ago
E<br> 3.6 What force is needed to give a mass of<br> 20 kg an acceleration of 5 m/s??
luda_lava [24]

Explanation:

  • Mass(m)= 20kg
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We know that,

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Hence, the needed force is 100N.

6 0
2 years ago
Read 2 more answers
During the time interval from 0.0 to 10.0 s, the position vector of a car on a road is given by x(t) = a + bt + ct2, with a = 17
Juli2301 [7.4K]

The car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

<h3>Average velocity of the car</h3>

The average velocity of the car is calculated as follows;

x(t) = a + bt + ct2

v = dx/dt

v(t) = b + 2ct

v(0) = -10.1 m/s + 2(1.1)(0) = -10.1 m/s

v(10) = -10.1 + 2(1.1)(10) = 11.9 m/s

<h3>Average velocity</h3>

V = ¹/₂[v(0) + v(10)]

V = ¹/₂ (-10.1  + 11.9 )

V = 0.9 m/s

Thus, the car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

Learn more about velocity here: brainly.com/question/4931057

#SPJ1

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