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2 years ago

If the atomic number of an atom is 4 and the mass number is 9, then how many electrons does it have

1 answer:

5 0

The atomic number of an atom says how many protons it has. This number cant change, since the atomic number is what gives elements their identities (in the periodic table, at least).
The mass number, on the other hand, says how many protons AND neutrons the atom has (so, the sum of P+ and N0). So, electrons have nothing to do with this number.
Atoms are neutrally charged, which means there has to be an equal number of positive and negative particles. The negative particles of an atom are its electrons, and since our atom has 4 protons, it must also have 4 electrons.

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Keeping in mind the kinetic energy of a moving vehicle, how can a driver best prepare to enter sharp curves in the roadway

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Gradually slow down to a safe speed well before entering the curve.

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2 years ago

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A ball is dropped off the side of a bridge.<br> After 1.55 s, how far has it fallen?

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2 years ago

A particular field is 111 yards long. Express this length in meters.

romanna [79]

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101.498 m

Explanation:

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3 years ago

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A block of a plastic material floats in water with 42.9% of its volume under water. What is the density of the block in kg/m3?

adell [148]

To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.

By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore

$\sum F = 0$

$F_b -W = 0$

$F_b = W$

$F_b = mg$

Here,

m = mass

g =Gravitational energy

The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore

$\rho_w V_{displaced} g = mg$

Remember the expression for which you can determine the relationship between mass, volume and density, in which

$\rho = \frac{m}{V} \rightarrow m = V\rho$

In this case the density would be that of the object, replacing

$\rho_w V_{displaced} g = V\rho g$

Since the displaced volume of water is 0.429 we will have to

$\rho_w (0.429V) = V \rho$

$0.429\rho_w= \rho$

The density of water under normal conditions is $1000kg / m ^ 3$ , so

$0.429(1000) = \rho$

$\rho = 429kg/m^3$

The density of the object is $429kg / m ^ 3$

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2 years ago

1. If an object of mass m collides and velocity v collides inelastically with an object of mass 3m that is initially at rest, th

Archy [21]

Answer:

1a). 4 times. 1b) 4. 1c) 1/4.

2a) 5 times. 1b) 5 1c) 1/5

3a) 7/3 times 3b) 7/3 3c) 3/7

4a) 8/5 times 4b) 8/5 4c) 5/8

Explanation:

1) Assuming no external forces acting during the collision, total momentum must be conserved, so the following general expression applies:

m₁*v₁₀ + m₂*v₂₀ = m₁*v₁f + m₂*v₂f (1)

If we assume that the collision is perfectly inelastic, this means that both masses stick together after the collision, so v₁f = v₂f.

If m₂ is initially at rest, ⇒ v₂₀ = 0.

Replacing in (1) we get the expression of vf as a function of v₁₀, as follows:

vf = v₁₀*(m₁/(m₁+m₂)

So, for the four cases we have the following:

1) initial mass = m

final mass = m+3m = 4 m

⇒final mass / initial mass = 4

vf = v₀* (m/4m) = v₀/4 ⇒v₀/vf = 4

So, the velocity of the system will decrease by a factor of 4. The new velocity will be vf= v₀/4.

2) Applying the same considerations, we get:

2a) final mass / initial mass = 5

2b) vf = v₀* (m/5m) = v₀/5 ⇒v₀/vf = 5

2c) vf = v₀/5

3) Applying the same considerations, we get:

3a) final mass / initial mass = 7/3

3b) vf = v₀* (3m/7m) =3/7* v₀ ⇒v₀/vf = 7/3

3c) vf = 3/7*v₀

4) Applying the same considerations, we get:

4a) final mass / initial mass = 8/5

4b) vf = v₀* (5m/8m) = 5/8*v₀ ⇒v₀/vf = 8/5

4c) vf = 5/8*v₀

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2 years ago