Answer:
The value is 
Explanation:
From the question we are told that
The mass of the object is 
The unstressed length of the string is 
The length of the spring when it is at equilibrium is 
The initial speed (maximum speed)of the spring when given a downward blow 
Generally the maximum speed of the spring is mathematically represented as

Here A is maximum height above the floor (i.e the maximum amplitude)
and
is the angular frequency which is mathematically represented as

So

=> 
Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

=> 
=> 
Generally at equilibrium position the net force acting on the spring is

=> 
=> 
So

=> 
Since they are travelling on two different straight
directions, therefore their paths will NEVER cross again. This because straight
lines will keep on going forever on the same direction therefore there is 0
possibility for two straight lines to paths cross again.
Answer:
<span>zero or never</span>
Stress, in physical sciences and engineering, force per unit area within materials that arises from externally applied forces, uneven heating, or permanent deformation and that permits an accurate description and prediction of elastic, plastic, and fluid behaviour.
Explanation:
hope i could help
<h3><u>Given </u><u>:</u><u>-</u></h3>
- The mass of the body is doubled
- The height of the body is constant
<h3><u>Solution </u><u>:</u><u>-</u><u> </u></h3>
We know that ,
Potential energy = mgh
<u>Therefore</u><u>, </u>
We can say that,
PE is directly proportional to Mass of the body
<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>
PE of the body = 2m * g * h. ...eq( I)
From (I) , we can conclude that, If mass of the body get doubled then its PE will also be doubled .
Answer:
230.4 N
Explanation:
From the question given above, the following data were obtained:
Charge (q) of each protons = 1.6×10¯¹⁹ C
Distance apart (r) = 1×10¯¹⁵ m
Force (F) =?
NOTE: Electric constant (K) = 9×10⁹ Nm²/C²
The force exerted can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²
F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰
F = 2.304×10¯²⁸ / 1×10¯³⁰
F = 230.4 N
Therefore, the force exerted is 230.4 N