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3 years ago

What is the best use of an atomic model to explain the charge of the particles in thomson's beams

Physics

1 answer:

hoa [83]3 years ago

6 0

The best use of an atomic model to explain the charge of the particles in Thomson's beams is:

An atom's smaller negative particles are at a distance from the central positive particles, so the negative particles are easier to remove.

Explanation:

In Thomson's model, an atom comprises of electrons that are surrounded by a group of positive particles to equal the electron's negative particles, like negatively charged “plums” that are surrounded by positively charged “pudding”.

Atoms are composed of a nucleus that consists of protons and neutrons . Electron was discovered by Sir J.J.Thomson. Atoms are neutral overall, therefore in Thomson’s ‘plum pudding model’:

atoms are spheres of positive charge
electrons are dotted around inside

Thomson's conclusions made him to propose the Rutherford model of the atom where the atom had a concentrated nucleus of positive charge and also large mass.

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What is the magnetic force on a proton that is moving at 5.2 x 107 m/s to the

alisha [4.7K]

Answer:

1.1648×10⁻¹¹ N

Explanation:

Using

F = qvBsinФ..................... Equation 1

Where F = Force on the proton, q = charge, v = velocity, B = magnetic Field, Ф = angle between the magnetic Field and the velocity.

Note: The angle between v and B = 90°

Given: v = 5.2×10⁷ m/s, B = 1.4 T, q = 1.6×10⁻¹⁹ C, Ф = 90°

Substitute into equation 1

F = 1.6×10⁻¹⁹(5.2×10⁷)(1.4)sin90°

F = 11.648×10⁻¹²

F = 1.1648×10⁻¹¹ N.

6 0

2 years ago

Read 2 more answers

A: How far did she travel? B: How long did she take?

allochka39001 [22]

A. 60 miles
B. 5 hours

Unless you are looking for slope, in which case the answer is different

7 0

2 years ago

A fire helicopter carries a 700-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee

rodikova [14]

Answer:

F = 50636.873 N

Explanation:

given,

bucket of water = 700-kg

length of cable = 20 m

Speed = 40 m/s

angle of the cable = 38.0°

let air resistance be = F

tension in rope be = T

T cos 38° = m×g..................(1)

$T sin 38^0= \dfrac{mv^2}{l} + F$ ..........(2)

equation (1)/(2)

$tan 38^0 =\dfrac{\dfrac{mv^2}{l} + F}{mg}$

$0.781=\dfrac{\dfrac{700\times 40^2}{20} + F}{700\times 9.8}$

F = 50636.873 N

Hence the force exerted on the bucket is equal to F = 50636.873 N

5 0

3 years ago

Which statement is the most appropriate conclusion for David to write?

ycow [4]

I need more to answer this

4 0

2 years ago

Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli

Masja [62]

Answer:

Not possible

Explanation:

$E_{cl}$ = longitudinal modulus of elasticity = 35 Gpa

$E_{ct}$ = transverse modulus of elasticity = 5.17 Gpa

$E_m$ = Epoxy modulus of elasticity = 3.4 Gpa

$V_{\rho l}$ = Volume fraction of fibre (longitudinal)

$V_{\rho t}$ = Volume fraction of fibre (transvers)

$E_f$ = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

$E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764$

Transverse modulus of elasticity is given by

$E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148$

$V_{\rho l}\neq V_{\rho t}$

Hence, it is not possible to produce a continuous and oriented aramid fiber.

5 0

3 years ago