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3 years ago

Why don't protons stick together

Physics

1 answer:

3241004551 [841]3 years ago

4 0

The strong nuclear force isnt quite strong enough to overcome electrosatic repulsion between two protons

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A 2.5kg object oscillates at the end of a vertically hanging light spring once every 0.65s .

sesenic [268]

Answer:

Explanation:

By the general expresion for this problem, we have:

y(t) = A*cos(w*t+∅)

since: w = 2π/T = 2π/0,65

For the initial conditions:

y(0) = (0.17cm)*cos(w*0+∅) = + 0.17 m ---> cos(∅) = 1 ---> ∅ = 0°

Then:

A) y(t) = (0.17 m)*cos((2π/0,65)*t)

B part

This means, find the first solution for:

y(t) = (0.17 m)*cos((2π/0,65)*t) = 0 > (equilibrium position)

then: cos((2π/0,65)*t) = 0 ---> (2π/0,65)*t = π/2 ---> t = 0.1625 sec

C part

By definition: (Velocity) v = dy/dt

Then, deriving: v = dy/dt = - (0.17 m)*(2π/0,65)*sin((2π/0,65)*t)

The maximum velocity ocurrs when sin((2π/0,65)*t) = ±1, then (in absolute value): Vmax = 1,64 m/s

D part

By definition: (Aceleration) A = dv/dt

Then, deriving: v = dv/dt = - (0.17 m)*(2π/0,65)²*cos((2π/0,65)*t)

The maximum aceleration ocurrs when cos((2π/0,65)*t) = ±1, then (in absolute value): Amax = 15,88 m/s²

E part

For the acceleration, ocurres by all the solution when:

cos((2π/0,65)*t) = cos(phase) = ±1, this means: phase = {π, 2π, 3π, ...}, all π multiples.

Then, for the position:

y(t) = (0.17 m)*±1= {+0.17 m ; -0.17 m}

For the velocity, ocurres by all the solution when:

sin((2π/0,65)*t) = sin(phase) = ±1, this means: phase = {π/2, π, 3π/2, ...}, all π/2 multiples.

Then, for the position:

y(t) = (0.17 m)*cos(phase)= 0

6 0

3 years ago

What does FCCLA stand for?

professor190 [17]

FCCLA Stands for Family, Career and Community Leaders of America.

6 0

3 years ago

Read 2 more answers

A weight of 200 n is hung from a spring with a spring constant of 2500 n/m and lowered slowly. How much will the spring stretch?

melisa1 [442]

The length by which the spring got stretched will be 0.08 m. The force is directly propotional to the distance by which the spring stretched.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =200

K is the spring constant= 2500 N/m

x is the length by which spring got stretched =?

The stretch of the spring is found as;

$\rm F=kx \\\\ x = \frac{F}{k} \\\\\ x= \frac{200}{2500} \\\\ x=0.08 \ m$