A physical change in something doesn't change what the it is. For example, if you break glass, it will still be glass. In a chemical change where there is a chemical reaction, a new thing is formed and energy is either given off or absorbed. For example, when you burn a log. The carbon in the log is reacting to the oxygen to create ashe and smoke
An object distance is
presented as s = 5f and we know that the mirror equation relates the image
distance to the object distance and the focal length.
The mirror equation is
1/f = 1/s + 1/s’ where the variable f stands for
the focal length of the mirror. Variable (s)
represents the distance between the mirror surface and the object and the
variable <span>(s’) represents the distance between the mirror surface and
the image. </span>
In addition, a concave mirror
will have a positive focal length (f) and a convex mirror will have a negative
focal length (f).
Now, we then have 1/f = 1/5f
+ 1/s’ which is s’ = 5f/4
Then we get the magnification
ratio that expresses the size or amount of magnification or reduction of the
object or image and to get the magnification, we use this equation: M= s’/s
M= 5f/4x5f
s’ = 1/4s
Therefore, the image height
is one fourth of the object height
Answer:
Stretch can be obtained using the Elastic potential energy formula.
The expression to find the stretch (x) is ![x=\sqrt{\frac{2\times EPE}{k}}](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B%5Cfrac%7B2%5Ctimes%20EPE%7D%7Bk%7D%7D)
Explanation:
Given:
Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.
To find: Elongation in the spring (x).
We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).
The formula to find EPE is given as:
![EPE=\frac{1}{2}kx^2](https://tex.z-dn.net/?f=EPE%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
Rewriting the above expression in terms of 'x', we get:
![x=\sqrt{\frac{2\times EPE}{k}}](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B%5Cfrac%7B2%5Ctimes%20EPE%7D%7Bk%7D%7D)
Example:
If EPE = 100 J and spring constant, k = 2 N/m.
Elongation or stretch is given as:
![x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B%5Cfrac%7B2%5Ctimes%20EPE%7D%7Bk%7D%7D%5C%5C%5C%5Cx%3D%5Csqrt%7B%5Cfrac%7B2%5Ctimes%20100%7D%7B2%7D%7D%5C%5C%5C%5Cx%3D%5Csqrt%7B100%7D%3D10%5C%20m)
Therefore, the stretch in the spring is 10 m.
So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.
These two forces are called action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects.
Hope this helps! :)
Answer:
<h2>1.17 m/s²</h2>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
![a = \frac{f}{m} \\](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7Bf%7D%7Bm%7D%20%20%5C%5C%20)
f is the force
m is the mass
From the question we have
![a = \frac{700}{600} = \frac{7}{6} \\ = 1.1666666...](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7B700%7D%7B600%7D%20%20%3D%20%20%5Cfrac%7B7%7D%7B6%7D%20%20%5C%5C%20%20%3D%201.1666666...)
We have the final answer as
<h3>1.17 m/s²</h3>
Hope this helps you