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rjkz [21]
3 years ago
11

Jay fills a wagon with sand (about 20 kg) and pulls it with a rope 40 m along the beach. He holds the rope 25â above the horizon

tal. The rope exerts a 20-N tension force on the wagon. How much work does the rope do on the wagon?
Physics
1 answer:
never [62]3 years ago
8 0

Answer:

W = 725 J

Explanation:

given,

mass of wagon = 20 kg

distance of pull = 40 m

angle made with the horizontal = 25°

tension force on the wagon = 20 N

Work done = ?

Horizontal component of the force will help in movement of the wagon.

Horizontal component of force=

F_x = F cos θ

Work done is equal to force into displacement

W = F.s

W =  F cos θ.s

W = 20 cos 25° x 40

W = 725 J

hence, work done on pulling the wagon is equal to W = 725 J

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A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.
bija089 [108]

a)

i) Potential for r < a: V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

ii) Potential for a < r < b:  V(r)=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

iii) Potential for r > b: V(r)=0

b) Potential difference between the two cylinders: V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c) Electric field between the two cylinders: E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

Explanation:

a)

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

E=\frac{\lambda}{2\pi \epsilon_0 r}

where

\lambda is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

E=0

So the potential where the electric field is zero is constant:

V=const.

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density +\lambda and an equal negative charge density -\lambda. Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)

However, we know that the potential at b is zero, so

V(r)=V(b)=0

ii) The electric field in the region a < r < b instead it is given only by the positive charge +\lambda distributed over the surface of the inner cylinder of radius a, therefore it is

E=\frac{\lambda}{2\pi r \epsilon_0}

And so the potential in this region is given by:

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r} (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

And so, for r<a,

V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

b)

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

- Potential at the surface of the outer cylinder:

V(b)=0

Therefore, the potential difference is simply equal to

V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c)

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

The electric field is just the derivative of the electric potential:

E=-\frac{dV}{dr}

so we can find it by integrating the expression for the electric potential. We find:

E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
A heat engine exhausts 8900 j of heat while performing 2800 j of useful work. part a what is the efficiency of this engine?
Anuta_ua [19.1K]
Given: Heat Qout means useful work = 2800 J

           Heat Qin = 8900 J

Required; Efficiency = ?

Formula: Efficiency = Qout/Qin = x 100%

                               = 2800 J/8900 J = 0.3146 X 100 %

               Efficiency = 31.46%


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Stronauts in space need to communicate with each other by radio links because
dybincka [34]

Answer:

Because there is nothing out in space , the sound waves from one astronaut's whistling can't travel over to the other astronaut's ears.

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When applying an arm sling, the injured arm's hand should be placed in a ________ position in the sling, and above the level of
Ierofanga [76]

Answer:

The answer is "thumb-up".

Explanation:

The sling often called the arm sling, is a solution for preventing either shoulder or elbow motion during healing. A triangle bandage could be constructed.

A sling is a strong tissue loop that descends from the head to hold the forearm. The armrests in the sling bent also at the elbow.

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