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Tomtit [17]
3 years ago
13

When do net force is applied to a moving object it still comes to rest because of its inertia

Physics
1 answer:
Paraphin [41]3 years ago
6 0
That is false for that.
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A block of mass m slides on a horizontal frictionless surface. The block is attached to a spring with a spring constant K. At th
Fittoniya [83]

Answer:

b) a = -k / m x , c) d²x / dt² = - A w² cos (wt+Ф) , d) and e)  T = 2π √m / k

h)   a = - A w² cos (wt+Ф)

Explanation:

a) see free body diagram in the attachment

b) We write Newton's second law

          Fe = m a

          -k x = ma

           a = -k / m x

c) the acceleration is

         a = d²x / dt²

     

      If x = A cos wt

        v = dx / dt = -A w sin (wt +Ф)

        a = d²x / dt² = - A w² cos (wt+Ф)

d) we substitute in Newton's second law

        d²x / dt² = -k / m x

   

We call

       w² = k / m

e) substitute to find w

     -A w² cos (wt+Ф) = -k / m A cos (wt+Ф)

      w² = k / m

Angular velocity and frequency are related

       w = 2π f

       f = 1 / T

       

 We substitute

      T = 2π / w

      T = 2π √m / k

g)    v= - A w sin (wt+Ф)

h) acceleration is

       a = - A w² cos (wt+Ф)

8 0
4 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
4 years ago
I WILL GIVE BRAINLIEST 50!!!!!
Inessa05 [86]
Transfer of heat through objects touching source
4 0
4 years ago
Read 2 more answers
Which spots in this picture show where a WARM air mass would form? *
Svet_ta [14]

Answer:

Spot B & Spot C

Explanation:

They're closer to the equator and get more direct solar radiation, making them more likely to be where a warm air mass would form.

3 0
3 years ago
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What do the spheres in the model represent
Sophie [7]

Answer: no

Explanation: we need a picture

4 0
3 years ago
Read 2 more answers
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