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Anon25 [30]
3 years ago
7

What is a small particle of an atom carries a neutral charge?

Physics
1 answer:
Sati [7]3 years ago
7 0
<em>It is found, with the positively charge protons in the central nucleus of the atom, while the negatively charges electrons rotate in orbits (Shells) around it.</em>

<em>Electron with a charge of -1</em>

<em>Electron- carries a negative energy </em>
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A sports car accelerates from rest at 4 metre per second squared for 10 seconds , its final velocity would be?
Sergeu [11.5K]

Answer:

40m/s

Explanation:

s= NOT NEEDED

u= 0

v= ?

a=4m/s^2

t=10

find the equation

v=u+at

substitute in values:

v=0+4x10

v=40m/s

6 0
3 years ago
Read 2 more answers
A 0.614mole sample of ideal gas at 12degree occupies a volume of 4.32.what is the pressure of the gas
Kaylis [27]

Answer:

336.9520 atm

Explanation:

The Gas Equation is as follows;-

Pressure×Volume=Number of Moles × Universal Gas Constant ×Temperature(in Kelvin)

Given Parameters

Number of moles-0.614 mol

Temperature 12°C or 12+273.15 ie 285.15°F

Volume-4.32 L

Universal Gas Constant-8.314 J/mol·K

Pressure -?(in atm)

Plugging in all the values in the Gas Equation:-

Pressure=\frac{0.614 × 8.314× 285.15}{4.32}   atm

Pressure=336.9520 atm

3 0
3 years ago
An aquarium has a volume of 2.75 cubic
Harlamova29_29 [7]
The aquarium can hold 2750 liters because 1 cubic meter equals 1000 liters.
3 0
2 years ago
Which of the following is the correct definition of light energy?
Sveta_85 [38]
B is the correct answer
3 0
2 years ago
Determine the gravitational potential energy, in kJ, of 3 m3 of liquid water at an elevation of 40 m above the surface of Earth.
melomori [17]

Explanation:

We will calculate the gravitational potential energy as follows.

                 P.E_{1} = mgz_{1}

       P.E_{1} = (\rho V)gz_{1}    

                    = 1000 kg/m^{3} \times 3 m^{3} \times 9.7 \times 40 m

                    = 1164000 J

or,                = 1164 kJ         (as 1 kJ = 1000 J)

Now, we will calculate the change in potential energy as follows.

             \Delta P.E = mg(z_{2} - z_{1})

                         = \rho \times V \times g (z_{2} - z_{1})

                         = 1000 \times 3 \times 9.7 (10 - 40)m

                         = -873000 J

or,                      = -873 kJ

Thus, we can conclude that change in  gravitational potential energy is -873 kJ.

4 0
2 years ago
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