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makvit [3.9K]
3 years ago
6

What did j.J. Thomson discover about the composition of atoms?

Physics
1 answer:
nordsb [41]3 years ago
5 0

Answer:

J.J. Thomson discovered <u><em>the electron</em></u> by experimenting with a Crookes, or cathode ray, tube. He demonstrated that cathode rays were negatively charged. In addition, he also studied positively charged particles in neon gas.

Explanation:

the answer is the underlined part of the answer section. I hoped this helped you a lot!! Study hard for whatever you are doing!!

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Find a unit vector normal to the plane containing Bold u equals 2 Bold i minus Bold j minus 3 Bold k and Bold v equals negative
Nastasia [14]

Answer:

\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}

Explanation:

It is given that,

\vec{u}=2i-j-3k

\vec{v}=-3i+j-2k

Taking the cross product of v and v such that,

\vec{w}=u\times v

\vec{w}=(2i-j-3k)\times (-3i+j-2k)

\vec{w}=5i+13j-k

|w|=\sqrt{5^2+13^2(-1)^2}

|w| = 13.92

Let \hat{w} is the unit vector normal to the plane containing u and v. So,

\hat{w}=\dfrac{\vec{w}}{|w|}

\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}

Hence, this is the required solution.

7 0
3 years ago
As the particles in a soup reach the bolling point what happens
ICE Princess25 [194]

Answer:

Once the particles of a liquid reach its boiling point, additional heat would cause the liquid particles to boil off or vaporize.

Explanation:

4 0
3 years ago
A glass rod and a steel rod are of equal length at 0C. At 100C they differ in length by
NeX [460]

The given lengths at 0 °C are 2.5 m

Let l₀ be the given lengths of the glass and steel rods at 0 °C. Let l and l' be the lengths of the glass and steel rods at 100 °C respectively.

From our expression for linear expansivity,

l = l₀ + l₀αΔθ where α = linear expansivity of glass = 0.000008/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Also,

l' = l₀ + l₀α'Δθ where α' = linear expansivity of steel = 0.000012/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Since the difference in their lengths at 100 °C = 0.001 m, we have that

l - l' = l₀ + l₀αΔθ - (l₀ + l₀α'Δθ)

l - l' = l₀ + l₀αΔθ - l₀ - l₀α'Δθ)

l - l' = l₀αΔθ - l₀α'Δθ

l - l' = l₀(α- α')Δθ

Making l₀ subject of the formula, we have

l₀ = (l - l')/[(α- α')Δθ]

Substituting the values of the variables into the equation, we have

l₀ = (l - l')/[(α- α')Δθ]

l₀ = 0.001 m/[(0.000008/°C - 0.000012/°C)100 °C.]

l₀ = 0.001 m/[(-0.000004/°C)100 °C.]

l₀ = 0.001 m/-0.0004

l₀ = -2.5 m

Neglecting the negative sign,

l₀ = 2.5 m

So, the given lengths at 0 °C are 2.5 m

Learn more about linear expansion here:

brainly.com/question/14089545

6 0
2 years ago
It takes 180 kj of work to accelerate a car from 21.0 m/s to 27.0 m/s. what is the car's mass?
beks73 [17]
558888544446556313321
4 0
3 years ago
A 25kg box fell 200m with an acceleration of 5 m/s2. with what force did it hit the floor when it landed?
Marizza181 [45]
According to Newton's Second Law of motion, the net force acting on the object is equal to its mass multiplied by its acceleration. In formula, it is written as

Net Force =mass * acceleration
Net force = 25 kg * 5m/s^2
Net force = 125 Newtons
6 0
3 years ago
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