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3 years ago

Compare the relative strengths of the nuclear force and the electric force

Physics

1 answer:

Serggg [28]3 years ago

6 0

Answer:

To establish this relationship we must examine the potentials that these forces create. The electrical potential is described by

Ve = k q / r

The potential for strong nuclear force is

Vn (r) = - gs / 4pir exp (-mrc / h)

Where gs is the stacking constant and r the distance between the nucleons,

We can compare these potentials where the force is derived from the relationship

E = -dU / dr

F = q E

Explanation:

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Suppose a rocket ship in deep space moves with constant acceleration equal to 9.80 m/s2, which gives the illusion of normal grav

DochEvi [55]

Answer:

a) 3673469.39 seconds

b) 6.61×10¹⁴ m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0.12×3×10⁸ m/s

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

Equation of motion

$v=u+at\\\Rightarrow 0.12\times 3\times 10^8=0+9.8t\\\Rightarrow t=\frac{0.12\times 3\times 10^8}{9.8}=3673469.39\ s$

Time taken to reach 12% of light speed is 3673469.39 seconds

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{(0.12\times 3\times 10^8)^2-0^2}{2\times 9.8}\\\Rightarrow s=6.61 \times 10^{14}\ m$

The distance it would have to travel is 6.61×10¹⁴ m

7 0

3 years ago

You wad up a piece of paper and throw it into the wastebasket. How far will

vitfil [10]

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, $g=9.8 m/s^2$ )

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

$d=\frac{u^2 sin 2\theta}{g}$

where

u is the initial velocity

$\theta$ is the angle of projection

g is the acceleration of gravity

For the piece of paper in this problem,

u = 4.3 m/s

$\theta=65^{\circ}$

Substituting,

$d=\frac{(4.3)^2 sin(2\cdot 65^{\circ})}{9.8}=1.45 m \sim 1.4 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0

3 years ago

Read 2 more answers

What are the two parts of a force pair?

iragen [17]

These two forces are called action and reaction forces and are the subject of Newton's third law of motion.

<em>Have a luvely day!</em>

8 0

3 years ago

PLEASE HELP!

SashulF [63]

Answer:

They gon' hate me regardless, that's why I do what I do (what I do)

See me in person i'm flawless (I'm flawless girl), i might just snatch up your dude (hahahaha)

Explanation:

i just want to be....appreciated

3 0

3 years ago

A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0

3 years ago