Answer with Explanation:
We are given that
Restoring force,
We have to find the work must you do to compress this spring 15 cm.
Using 1 m=100 cm
Work done=
W=
![W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}](https://tex.z-dn.net/?f=W%3Dk%5B%5Cfrac%7B%28%5CDelta%20s%29%5E2%7D%7B2%7D%5D%5E%7B0.15%7D_%7B0%7D%2Bq%5B%5Cfrac%7B%28%5CDelta%20s%29%5E4%7D%7B4%7D%5D%5E%7B0.15%7D_%7B0%7D)
Ideal spring work=
Percentage increase in work=
%
Answer:
y and length is directly relation
Explanation:
Given data
A single-slit diffraction pattern is formed on a distant scree
angles involved = small
to find out
what factor will the width of the central bright spot on the screen change
solution
we know that for single slit screen formula is
mass ƛ /area = sin θ and y/L = sinθ
so we can say mass ƛ /area = y/L
and y = mass length ƛ / area .................1
in equation 1 here we can see y and length is directly relation so we can say from equation 1 that the width of the central bright spot on the screen change if the distance from the slit to the screen is doubled
In physics<span>, a </span>force<span> is said to do </span>work<span> if, when acting, there is a </span>displacement<span> of the point of application in the direction of the force. For example, when a ball is held above the ground and then dropped, the work done on the ball as it falls is equal to the weight of the ball (a force) multiplied by the distance to the ground (a displacement).
power= work over time
the more power you put into your work the less time it will take, if you have less power the more time</span>
R = ρl/A
From the equation above R = Resistance, l = length, A = Cross Sectional Area of wire.
From the equation, it can be seen that R would increase if the wire's area is reduced.
If the area of the wire is reduced, means the same thing as:
<span>A. decreasing the wire’s thickness</span>
