To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as

Here,
k = Coulomb's constant
q = Charge of proton and electron
r = Distance
Replacing we have that,


The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.
The acceleration of the electron is given as



The acceleration of the proton is given as,



Answer:
-0.7 m/sec
Explanation:
Mass of first block = m1 =3.0 kg
Mass of second block = m2= 5.0 kg
Velocity of first block = V1= 1.2 m/s
Velocity of second block = V2 = ?
Momentum of Center of mass MVcom is sum of both blocks momentum and is given by
MVcom= m1v1+m2v2
Where
M= mass of center of mass
Vcom= Velocity of center of mass=0 m/s (because center of mass is at rest , so Vcom = 0 m.sec)
Putting values, we get;
0= 3×1.2+5v2
==> v2= 3.6/5= - 0.7 m/s
-ve sign indicates that block 2 is moving in opposite direction of block 1
First, when the student added the layers of wax over each other, this became a representation of sedimentary rocks.
Then the student folded his/her palm and squeezed the layers of wax. This means that the student applied heat and pressure on the wax (sedimentary rocks)
Referring to the diagram below which represents the rock cycle, we will find that applying heat and pressure on sedimentary rocks would convert these rocks into metamorphic rocks.
Based on the above, the best choice would be:<span>d. Heat and pressure can change sedimentary rocks into metamorphic rocks.</span>
True
False
True
My answers
Answer:
a) t=24s
b) number of oscillations= 11
Explanation:
In case of a damped simple harmonic oscillator the equation of motion is
m(d²x/dt²)+b(dx/dt)+kx=0
Therefore on solving the above differential equation we get,
x(t)=A₀
where A(t)=A₀
A₀ is the amplitude at t=0 and
is the angular frequency of damped SHM, which is given by,

Now coming to the problem,
Given: m=1.2 kg
k=9.8 N/m
b=210 g/s= 0.21 kg/s
A₀=13 cm
a) A(t)=A₀/8
⇒A₀
=A₀/8
⇒
applying logarithm on both sides
⇒
⇒
substituting the values

b) 

, where
is time period of damped SHM
⇒
let
be number of oscillations made
then, 
⇒