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3 years ago

Which of the following is incorrect for nuclear forces

Physics

2 answers:

densk [106]3 years ago

4 0

Answer:

c they obey inverse square law

bija089 [108]3 years ago

4 0

Answer is C .

The reason is as follows:
Nuclear forces don’t obey inverse square law.

Neither the strong nuclear force nor the weak nuclear force follow the inverse square law.

The weak nuclear force is a Yukawa-type force. Such a force is characterized by a potential in the form ∝−/
∝
e
−
m
r
/
r
. At short range ( ≪1
m
r
≪
1
) the exponential term is just 1
1
, so the potential becomes an inverse-
r
potential with an inverse square force law associated with it. So at very short (subatomic) ranges, the weak nuclear force is actually very similar to the electromagnetic force. But as soon as >1
m
r
>
1
, the force rapidly vanishes and becomes undetectable.

The strong nuclear force does not decrease with distance. Which means that as you “stretch” the bond between two quarks, for instance, more and more energy needs to be invested; until there is enough energy to create a new quark-antiquark pair, at which point the bond breaks. This limits the range of the strong nuclear force to subatomic scales, too.

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What force acts in the opposite direction to an object moving through the air?

evablogger [386]

Air resitances also know as drag is a force that is caused by air the force acts in oppsite directions to an object moving through the air ..

HOPE THIS HELP YOU ..

5 0

3 years ago

Two protons are 9.261 fm apart. (1 fm= 1 femtometer = 1 x 10^-15 m.) What is the ratio of the electric force to the gravitationa

Strike441 [17]

Answer:

$1.24\times 10^{36}$

Explanation:

$q$ = magnitude of charge on each proton = 1.6 x 10⁻¹⁹ C

$m$ = mass of each proton = 1.67 x 10⁻²⁷ kg

r = distance between the two protons = 1 x 10⁻¹⁵ m

Electric force between the two protons is given as

$F_{e} = \frac{kq^{2}}{r^{2}}$

$F_{e} = \frac{(9\times 10^{9})(1.6\times 10^{-19})^{2}}{(1\times 10^{-15})^{2}}$

$F_{e} = 230.4$ N

Gravitational force between the two protons is given as

$F_{g} = \frac{Gm^{2}}{r^{2}}$

$F_{g} = \frac{(6.67\times 10^{-11})(1.67\times 10^{-27})^{2}}{(1\times 10^{-15})^{2}}$

$F_{g} = 1.86\times 10^{-34}$ N

Ratio is given as

$Ratio =\frac{F_{e}}{F_{g}}$

$Ratio =\frac{230.4}{1.86\times 10^{-34}}$

$Ratio = 1.24\times 10^{36}$

3 0

3 years ago

Read 2 more answers

hallar un numero de dos cifras sabiendo que la suma de las cifras es 12 y que la primera de ellas es el triple de la segunda

Natali [406]

Answer:

93 o 39.

Explanation:

Hola.

En este caso, podemos resolver este problema por medio del planteamiento de las siguientes ecuaciones:

1. Definimos x como el primer dígito y y como el segundo, por lo tanto, como su suma es 12, escribimos:

x + y = 12

2. Ahora, dado que el primer dígito, x es el triple del segundo, escribimos:

x = 3 y

De este modo, podemos resolver para y:

3 y + y = 12

4 y = 12

y = 12 / 4

y = 3

Y para x:

x = 3 y

x = 3 * 3

x = 9

Así, el número puede ser 93 o 39.

¡Saludos!

3 0

3 years ago

An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up. Now a slab of dielectric material is

natita [175]

Answer:

The charge on the capacitor had increased

Explanation:

The expression for the capacitance of an air-filled parallel-plate capacitor is as follows as;

$C=\frac{\epsilon _{0}A}{d}$

Here, C is the capacitance, $\epsilon _{0}$ is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.

When a slab of dielectric material is placed between the plates of the capacitor then the expression for the capacitance is as follows;

$C=\frac{K\epsilon _{0}A}{d}$

Here, K is the dielectric constant.

In the given problem, a slab is inserted between the plates of the capacitor then the capacitance of the capacitor will increase in this case. Therefore, the option (a) is true.

The expression for the charge stored in the capacitor is as;

Q= CV

Here, Q is the charge and V is the potential.

The charge will also increase in this case as the charge stored in the capacitor is directly proportional to the capacitance. Therefore, the option (d) is not true.

The expression for the energy stored in the capacitor is as follows;

$E=\frac{CV^{2}}{2}$

The voltage is constant in the given problem but the capacitance increases then the energy stored in the capacitor will increase. Therefore, the option (b) is not true.

The voltage across the capacitor will remain same as the capacitor is still connected to the battery. Therefore, the option (c) is not true.

Therefore, only option (a) is true.

3 0

3 years ago

In need of some help here

Cloud [144]

Answer:

need more context!

Explanation:

7 0

3 years ago