Explanation:
Given that,
Work done to stretch the spring, W = 130 J
Distance, x = 0.1 m
(a) We know that work done in stretching the spring is as follows :

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m
So,

So, the new work is more than 130 J.
I answered the question but it got deleted?? why?
Weight is equivalent to the product of the mass of an object and the strength of the gravitational field.
Using:
F = ma
a = 8.2 / 5
a = 1.64 N/kg
The gravitational field strength is equivalent to 1.64 N/kg.
High amplitude sound would be music, radio, or earthquakes.
Low amplitude sound would be a breeze or wind.
Answer:
The diameter is EF
Explanation:
Given
Circle A (See attachment)
Required
Determine the line that represents the diameter
First, it should be noted that the diameter of a circle is always a straight line.
From the attachment, the circle has the following straight lines:
It should also be noted that the diameter passes through the center of a circle and divides it into two congruent parts.
From the list of straight lines above, only line EF satisfy this property
Hence, the diameter of the circle is line EF.