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2 years ago

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A train is travelling at speed of 20m/s. Brakes are applied so as to produce a uniform acceleration of -0.5m/s2. find how far th

e train will go before it is brought to rest.

1 answer:

aivan3 [116]2 years ago

7 0

Use v^2 - u^2 = 2as

Here, u = 20 m/s , a = -0.5 m/s^2 and v = 0

- 400 = -s ======> s = 400 m

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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.

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7 0

3 years ago

a) ¿En qué posición es mínima la magnitud de la fuerza sobre la masa de un sistema masa-resorte? 1) x 0, 2) x A o 3) x A. ¿Por q

Tatiana [17]

Answer:

a) the correct answer is 1 , b) x=0 F=0, a=0

x= 0.050 F= -7.5 N, a= -15 m/s²

x= 0.150 F= 22.5 N, a=- 45 m/s²

Explanation:

a) In a mass - spring system the force is given by the Hooke force,

F = - k x

Analyzing this equation we see that the outside is proportional to the elongation from the equilibrium position, therefore the force is zero when the spring is in its equilibrium position

the correct answer is 1

b) we assume that the given values are from the equilibrium position of the spring.

Let's calculate the force

x = 0

F = 0

x = 0.050

F = - 150 0.050

F = - 7.50 N

x = 0.150

F = - 150 0.150

F = - 22.5 N

let's use Newton's second law to find the acceleration

F = m a

a = F / m

x = 0 m

a = 0

x = 0.050 m

a = -7.50 / 0.50

a = - 15 m / s²

x = 0.150 m

a = - 22.5 / 0.50

a = - 45 m/s²

TRASLATE

a) En un sistema masa – resorte la fuerza es dada por la fuerza de Hoke,

F= - k x

analizando esta ecuación vemos que la fuera es proporcional a la elongación desde la posición de equilibrio, por lo tanto la fuerza es cero cuando el resorte esta en su posición de equilibrio

la respuesta correcta es 1

b)suponemos que los valores dados son desde la posición de equilibrio del resorte.

Calculemos la fuerza

x=0

F= 0

x=0.050

F = - 150 0.050

F= - 7.50 N

x= 0.150

F= - 150 0.150

F= - 22.5 N

usemos la segunda ley de Newton para encontrar la aceleración

F = m a

a = F/m

x =0 m

a = 0

x= 0.050 m

a = -7.50/ 0.50

a =- 15 m/s²

x= 0.150 m

a= - 22.5 / 0.50

a= - 45 m/s²

7 0

3 years ago

If there is no electric current flowing through a wire then what happens to the magnetic field

Zinaida [17]

If there is no current in the wire .....the direction of magnetic field remains unchanged

7 0

2 years ago

Read 2 more answers

a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?

Verdich [7]

The distance between the two charges is $0.3 m$

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges of the two objects

r is the separation between the two charges

In this problem, we are given the following:

$q_1 = 8\cdot 10^{-5} C$

$q_2 = 3\cdot 10^{-5} C$

$F=2.4\cdot 10^2 N$

Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

$r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m$

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3 years ago

Iodine molecules separates into atoms after absorbing light of 450 nm. If one

OlgaM077 [116]

Answer:E = hc/? = 4.41 x 10-19 J

Energy absorbed by each atom :

E (atom) = 2.205 x 10-19 J

Now Bond Energy of each molecule (B) = 3.98 x J

So, for each atom 1.99 x 10-19 J

So now

KE of each atom = E(atom) - B (atom)

= 2.15 x 10-19 J

5 0

3 years ago