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3 years ago

What happens if 5 Protons combine with 4 neutrons.

2 answers:

7 0

in the case of beryllium, a light metal, only the beryllium-9 isotope is stable with its 9 nucleons (i.e. 4 protons and 5 neutrons)

Hope this helps so sorry if it does not help

JulijaS [17]3 years ago

6 0

They go looking for 5 electrons to join them and make one whole atom of Boron.

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Three tiny charged metal balls are arranged on a straight line. The middle ball is positively charged and the two outside balls

Dmitrij [34]

Answer:

$(a) 189.23 N$ , $(b) 47.31 N$ and $(c) 141.92 N$ .

Explanation:

Three balls are shown in figure having charge $q=1.45 \mu C$ . The middle ball, $B$ , is positively charged having charge $+q$ , and the remaining two outside balls, $A$ and $C$ , are negatively charged having charged $-q$ as shown.

$AC=20 cm$ and $AB=BC=10$ cm as B is the mid-point of AC.

Let $d_1=AC=20\times 10^{-3}m$ and $d_2=AB=BC=10\times 10^{-3}m$

From Coulomb's law, the magnitude of the force, $F$ , between two point charges having magnitudes $q_1 \& q_2$ , separated by distance, $d$ , is

$F=\frac {1}{4\pi\epsilon_0}\frac {q_1q_2}{d^2}\;\cdots (i)$

where, $\epsilon_0$ is the permittivity of free space and

$\frac {1}{4\pi\epsilon_0}=9\times 10^9$ in SI units.

This force is repulsive for the same nature of charges and attractive for the different nature of charges.

Now, Using equations(i),

(a) The magnitude of attraction force between balls A and B is

$F_{AB}=F_{BC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_2)^2}$

$\Rightarrow F_{AB}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(10\times 10^{-3}\right)^2}$

$\Rightarrow F_{AB}=189.23 N$

(a) The magnitude of the repulsive force between balls A and C is

$F_{AC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_1)^2}$

$\Rightarrow F_{AC}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(20\times 10^{-3}\right)^2}$

$\Rightarrow F_{AC}=47.31 N$

$F_{net}=189.23-47.31 N$

$\Rightarrow F_{net}=141.92 N$

4 0

3 years ago

Four distinguishable particles move freely in a room divided into octants (there are no actual partitions). Let the basic states

mafiozo [28]

Answer:

Explanation:

Since the door that leads to the room is opened, this gives room for particles to move into the next identical room and divided into octants. Now the amount of space that can be occupied becomes double, the number of basic states has increased by 404916

8 0

3 years ago

A bullet has a mass of 8 grams and a muzzle velocity of 340m/sec. A baseball has a mass of 0.2kg and is thrown by the pitcher at

Thepotemich [5.8K]

Answer:

Momentum of bullet

$P = 2.72 kg m/s$

momentum of baseball

$P = 8 kg m/s$

Explanation:

As we know that momentum is defined as the product of mass and velocity

here we know that

mass of the bullet = 8 gram

velocity of bullet = 340 m/s

momentum of the bullet is given as

$P = mv$

$P = (\frac{8}{1000})(340)$

$P = 2.72 kg m/s$

Now we have

mass of baseball = 0.2 kg

velocity of baseball = 40 m/s[/tex]

momentum of baseball is given as

$P = (0.2)(40)$

$P = 8 kg m/s$

3 0

4 years ago

You nose out another runner to win the 100.000m dash. If your total time for the race was 11.8 seconds and you aced out the othe

daser333 [38]

You completed 100.000 m in 11.800 second

you aced out the other runner by 0.001 seconds

time taken by other runner to finish = 11.800 + 0.001 = 11.801s

speed of other runner = 100/11.801 m/s

distance covered by him in 11.8 s = 100/11.801 × 11.8 = 99.992 m

you won by = 100 - 99.992 = 0.008 meter

3 0

4 years ago

A freight train consists of two 8.00 \times 10^5~\text{kg}8.00×10 5 kg engines and 45 cars with average masses of 5.50 \time

erma4kov [3.2K]

Answer:

F₁ = 4.29 x 10⁵ N

Explanation:

The total force required to move the freight train with the given acceleration is given by the following formula:

F = ma + f

where,

F = Total Force Required from both engines = ?

m = equivalent mass of system = 2(8 x 10⁵ kg) + 5.5 x 10⁵ kg = 21.5 x 10⁵ kg

a = required acceleration = 5 x 10⁻² m/s²

f = force of friction = 7.5 x 10⁵ N

Therefore,

F = (21.5 x 10⁵ kg)(0.05 m/s²) + 7.5 x 10⁵ N

F = 8.575 x 10⁵ N

Now, for identical forces in each engine can be given as:

Force exerted by each engine = F₁ = F/2

F₁ = 8.575 x 10⁵ N/2

3 0

3 years ago