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Masteriza [31]
1 year ago
8

The charged particles in the beams that Thomson studied came from atoms. As these particles moved away from their original atoms

, they formed a visible beam. The current model of the atom includes protons, neutrons, and electrons.
Physics
2 answers:
iris [78.8K]1 year ago
6 0

The current model of atom that includes protons, neutrons, and electrons is the model of Bohr. In his atomic model, the proton and neutrons are inside the nucleus and that the electrons orbit around the dense nucleus forming a visible beam.

aalyn [17]1 year ago
5 0

The question to the above information is;

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

Answer;

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

Explanation;

-Atoms are comprised of a nucleus consisting of protons (red) and neutrons (blue). The number of orbiting electrons is the same as the number of protons and is termed the "atomic number" of the element.

J.J. Thomson discovered the electron. Atoms are neutral overall, therefore in Thomson’s ‘plum pudding model’:

  • atoms are spheres of positive charge
  • electrons are dotted around inside
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A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the
lions [1.4K]

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

3 0
2 years ago
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
4 0
1 year ago
In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
scoray [572]

Answer:

W = 506.75 N

Explanation:

tension = 2300 N

Rider is towed at a constant speed means there no net force acting on the rider.

hence taking all the horizontal force and vertical force in consideration.

net horizontal  force:

F cos 30° - T cos 19° = 0

F cos 30° = 2300 × cos 19°

F = 2511.12 N

net vertical force:

F sin 30° - T sin 19°- W = 0

W = F sin 30° - T sin 19°

W =  2511.12 sin 30° - 2300 sin 19°

W = 506.75 N

8 0
2 years ago
Which statements can be inferred from the Paleozoic era time scale? There was volcanic activity during the Paleozoic era. Dinosa
Alchen [17]

Answer:

A,C,E

Explanation:

5 0
1 year ago
Read 2 more answers
Ocean waves are observed to travel to the right along the water surface during a developing storm. A Coast Guard weather station
Nuetrik [128]

Answer:

The amplitude is  2.3 m

The Wavelength is 8.6 m

The frequency is 0.16 Hz

The time period is 6.25 sec

The equation that governs the behavior is  Y=(2.3)sin[(\frac{2\pi}{8.6} )x -(\frac{2\pi}{6.2} )t]

Explanation:

The explanation is shown on the first uploaded image

6 0
2 years ago
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