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Harrizon [31]
3 years ago
10

At what distance is the electrostatic force between two protons equal to the weight of one proton?

Physics
1 answer:
bearhunter [10]3 years ago
5 0

Answer:

The distance between two protons is 37 cm.                    

Explanation:

The electric force between two electric charges is given by :

F=k\dfrac{q_1q_2}{d^2}

Here,

k is the electrostatic force

d is the separation between charges

q_1\ and\ q_2 are charges

According to given condition, the electrostatic force between two protons equal to the weight of one proton. So,

W_p=\dfrac{kq^2}{d^2}

d=\sqrt{\dfrac{kq^2}{W_p}}

As protons are very small particle, W = m

d=\sqrt{\dfrac{kq^2}{m}}

d=\sqrt{\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(1.67\times 10^{-27})}}

d = 0.37 m

or

d = 37 cm

So, the distance between two protons is 37 cm. Hence, this is the required solution.

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If the atomic mass of Sodium-18 is 18.02597 u, what is the binding energy?
Klio2033 [76]

Answer:

<h3>The binding energy of sodium Na=<em>5.407791×10⁹J</em></h3>

Explanation:

<h3>Greetings !</h3>

Binding energy, amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Binding energy is especially applicable to subatomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals.

<h2>Formula : Eb=(Δm)c²</h2><h3>where:Eb= binding energy</h3><h3> .Δm= mass defect(kg)</h3><h3> c= speed of light 3.00×10⁸ms¯¹</h3><h2 /><h3><u>Given</u><u> </u><u>values</u></h3>
  • m= 18.02597
  • c=3.00×10⁸ms¯¹

<h3><u>required </u><u>value</u></h3>
  • Eb=?

<h3><u>Solution:</u></h3>
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1 year ago
Bob has a brother, jim, who has a daughter named bertha. Bertha's daughter, jennifer, has a sister named penny. which of the fol
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A 50.0 kg object rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and
MariettaO [177]

Answer:

f=140\ N

Explanation:

Given:

  • mass of the object on a horizontal surface, m=50\ kg
  • coefficient of static friction, \mu_s=0.3
  • coefficient of kinetic friction, \mu_k=0.2
  • horizontal force on the object, F=140\ N

<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>

F_s=\mu_s.N

where:

N= normal force of reaction acting on the body= weight of the body

F_s=0.3\times (50\times 9.8)

F_s=147\ N

As we know that the frictional force acting on the body is always in the opposite direction:

So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.

so, the frictional force will be:

f=140\ N

8 0
3 years ago
You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The
Sliva [168]

Answer:

Explanation:

Given

Height of ceiling is h=3.6\ m

Initial speed of Putty u=9.5\ m/s

Speed of Putty just before it strike the ceiling is given by

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v^2-9.5^2=2\times (-9.8)\times 3.6

v^2=19.69

v=4.43\ m/s

time taken by putty to reach the ceiling

v=u+at

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t=\frac{5.07}{9.8}

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3 years ago
To stretch a certain nonlinear spring by an amount x requires a force F given by F = 40 x − 6 x 2 , where F is in Newtons and x
strojnjashka [21]

Answer:

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Explanation:

The potential energy change of the spring ∆U = -W where W = work done by force, F.

Now W = ∫F.dx

So, ∆U = - ∫F.dx = - ∫Fdxcos180 (since the spring force and extension are in opposite directions)

∆U = - ∫-Fdx

=  ∫F.dx

Since F = 40x - 6x² and x moves from x = 0 to x = 2 m, we integrate thus, ∆U =  ∫₀²F.dx

=  ∫₀²(40x - 6x²).dx

=  ∫₀²(40xdx - 6x²dx)

=  ∫₀²(40x²/2 - 6x³/3)

=  ∫₀²(20x² - 2x³)

= [20x² - 2x³]₀²

= [(20(2)² - 2(2)³) - (20(0)² - 2(0)³)

= [(20(4) - 2(8)) - (0 - 0))

= [80 - 16 - 0]

= 64 J

5 0
3 years ago
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