Answer:
The period of the resulting oscillatory motion is 0.20 s.
Explanation:
Given that,
Spring constant 
We need to calculate the time period
The object is at rest and has no elastic potential but it does has gravitational potential.
If the object falls then the the gravitational potential change in to the elastic potential.
So,
Where,h = distance
k = spring constant
Put the value into the formula
Using formula of time period
Put the value into the formula
Hence, The period of the resulting oscillatory motion is 0.20 s.
The velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
<h3>What is Electric field?</h3>
Electric field is the physical field that surrounds a charge.
<h3>How to find final velocity of the electron when it moves some distance in a certain electric field?</h3>
- From Newton's second law, the acceleration the electron will be
a=F/m=qE/m
- where q= charge of electron
E= electric field
m= mass of electron
=(−1.60×10^−19C)(3×10³N/C)/(9.11×10^-31kg)
=10¹⁵×0.526m/s²
- The kinematics equation v²=v0²+2a(Δx)
- where v=final velocity of the electron
v0=initial velocity of the electron =5×10⁶m/s
a=acceleration of the electron =10¹⁵×0.526m/s²
Δx=distance moved by the electron in east direction =1cm=10^-2m
- Now v^2=(5×10⁶)²+2×10¹⁵×0.526×10^-2
=25×10¹²+10.52×10¹²
=35.52×10¹²
- Now velocity of electron=5.95×10⁶m/s.
Thus , we can conclude that the velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.
Learn more about electric field here:
brainly.com/question/26199225
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Answer is C.
Answer: 100lbs
Explanation: The Earth pushes you down at 100lbs, so you push down the earth by 100lbs which is enough to keep you firmly attached to the ground and not allow you to jump more than a couple of feet.
