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3 years ago

Categorize the particles according to whether they are found in the nucleus or the electron cloud.

Physics

2 answers:

Ray Of Light [21]3 years ago

5 0

Answer:

Nucleus: Neutral particles, positively charged particles

Electron Cloud: Negatively charged particles

Explanation:

inna [77]3 years ago

4 0

1. neutral particles (neutrons) are in the nucleus
2. nucleus is in the nucleus
3. electron cloud is in the electron cloud
4. positively charged particles (protons) are in the nucleus
5. negatively charged particles (electrons) are in the electron cloud

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The answer is A bc I did the quiz and I got it right

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3 years ago

A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The d

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The average power needed to produce this final speed is 1069.1 hp.

Mass of the dragster, m = 500.0 kg,

Displacement travelled by the dragster, s = 402 m,

Time taken in this travel, t = 5.0 s,

Final velocity of the dragster, v = 130 m/s.

Let the initial velocity of the dragster be u and acceleration be a.

Using kinematical equation, s = ut + (1/2)at^2.

402 = u*5 + (1/2)*a*5^2

10*u + 25*a = 804. ...........(1)

Using kinematical equation, v = u +at.

130 = u + 5*a

5*u + 25*a = 650. .............(2)

Solving (1)and (2), we get,

u = 30.8 m/s.

According to work-energy theorem,

Work done = change in kinetic energy

W = (1/2)*m*(v^2 - u^2)

W = (1/2)*500*(130^2 - 30.8^2)

W = 3987840. J

Therefore power rating of the dragster is given by,

P ⇒ W/t. = 3987840/5 = 797568 watt.

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Learn more about Power rating here brainly.com/question/20137708

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2 years ago

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A box with a mass of 7 kilograms is pushed up a ramp to a height of 5 meters. What is the work done against the force of gravity

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3 years ago

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A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

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Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

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3 years ago

Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are tripled and their separation

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Answer:

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Explanation:

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2 years ago