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AnnyKZ [126]
3 years ago
6

Describe the structure of an atom.

Physics
1 answer:
Maurinko [17]3 years ago
5 0
A as an atom contains a nucleus where protons and neutrons are present. Electrons orbit the nucleus in paths called orbital
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When the Moon orbits Earth, what is the centripetal force?
nata0808 [166]

Answer:

Gravity is the centripetal force when the moon orbits the earth.

5 0
2 years ago
A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp
Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m =  0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2mU^{2}

K.E = 1/2 x 0.17 x 6^{2}

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/s^{2}

To calculate the final velocity, let us use third equation of motion.

V^{2} = U^{2} + 2as

V^{2}  = 6^{2} + 2 x 0.3 x 61

V^{2} = 36 + 36

V^{2} = 72

V = \sqrt{72}

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

5 0
2 years ago
1 example of a conductor and 1 example of a insulator in your EVERYDAY world.
ratelena [41]

Answer: Examples of conductors include metals, aqueous solutions of salts (i.e., ionic compounds dissolved in water), graphite, and the human body. Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.

4 0
2 years ago
A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the amplitude of the m
Anna35 [415]

To solve this problem it is necessary to take into account the concepts of Intensity as a function of Power and the definition of magnetic field.

The intensity depending on the power is defined as

I = \frac{P}{4\pi r^2},

Where

P = Power

r = Radius

Replacing the values that we have,

I = \frac{60}{(4*\pi (0.7)^2)}

I = 9.75 Watt/m^2

The definition of intensity tells us that,

I = \frac{1}{2}\frac{B_o^2 c}{\mu}

Where,

B_0 =Magnetic field

\mu = Permeability constant

c = Speed velocity

Then replacing with our values we have,

9.75 = \frac{Bo^2 (3*10^8)}{(4\pi*10^{-7})}

Re-arrange to find the magnetic Field B_0

B_o = 2.86*10^{-7} T

Therefore the amplitude of the magnetic field of this light is B_o = 2.86*10^{-7} T

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3 years ago
What is the working principle of an electric motor?
slava [35]
The answer B homie trust
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3 years ago
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