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2 years ago

Which scientist invented a model of the atom that most closely resembles the modern electron cloud model

Physics

1 answer:

Alex Ar [27]2 years ago

4 0

Bohr, he invented the Bohr model which is the basis for the beginning of quantum physics.

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3. What do we call the ONLY part of the electromagnetic spectrum that we can

otez555 [7]

Answer:

Visible Light

wavelength = 4000 - 7000 Angstroms = 400 - 700 milli-microns

1 A unit = 10^-10 m

1 mμ = 10^-9 m

6 0

2 years ago

The number of sound waves per unit time is called what

Andreyy89

Answer:

Frequency

Explanation:

Hoped this helped

3 0

3 years ago

8. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott

nasty-shy [4]

Answer: The ball is travelling with a speed of 5.5 m/s after hitting the bottle.

Explanation:

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

$m_1,u_1\text{ and }v_1$ are the mass, initial velocity and final velocity of ball.

$m_2,u_2\text{ and }v_2$ are the mass, initial velocity and final velocity of bottle.

We are given:

$m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s$

Putting values in above equation, we get:

$(0.4\times 18)+(0.2\times 0)=(0.4\times v_1)+(0.2\times 25)\\\\v_1=5.5m/s$

Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

5 0

3 years ago

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]

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:

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):

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